//这是模拟的代码，因为用了二维数组，空间开不了多大，会提示“too large”
//开小了又会RE，这里只是用来辅助大家理解题意。
#include <cstdio>
#include <iostream>
using namespace std;
int n, a, b, g, k, x, y;
int map[7010][7010];
template<class T>inline void read(T &res)
{
    static char ch;
    while( (ch=getchar()) < \'0\' || ch > \'9\');
        res = ch - 48;
    while( (ch = getchar() ) >= \'0\' && ch <= \'9\')
        res = ch - 48 + res * 10;
}
int main()
{
    freopen("carpet.in", "r", stdin);
    freopen("carpet.out", "w", stdout);
    read(n);
    for (int i = 1; i <= n; i++)
    {
        read(a);read(b);read(g);read(k);
        for (int l = b; l <= b+k; l++)
            for (int j = a; j <= a+g; j++)
                map[l][j] = i;
    }
    read(x); read(y);
    if (map[y][x] != 0) cout << map[y][x];
    else cout << -1;
    return 0;
}

//第二种直接判断的代码，时空复杂度都相当优。 
#include <stdio.h>
#include <iostream>
#include <stdlib.h>
using namespace std;
int i, j, k, n, x, y;
int ans[10005][5] = {0};
template<class T>inline void read(T &res)
{
    static char ch;
    while( (ch=getchar()) < \'0\' || ch > \'9\');
        res = ch - 48;
    while( (ch = getchar() ) >= \'0\' && ch <= \'9\')
        res = ch - 48 + res * 10;
}
int main()
{
    read(n);
    for(i = 1; i <= n; i++)
        for(j = 1; j <= 4; j++)
            read(ans[i][j]);
    read(x);read(y);
    for(i = n; i >= 1; i--)
        if( ( (ans[i][1] + ans[i][3] ) >= x) && ( (ans[i][2] + ans[i][4] ) >= y ) && ans[i][1] <= x && ans[i][2] <= y )
        {
            cout << i;
            return 0;
        }
    cout << -1;
    return 0;
}

#include <cstdio>
#include <iostream>
using namespace std;
int n, k, p, ans;
int t = 1;
int color[200010];
int house[200010];
template<class T>inline void read(T &res)
{
    static char ch;
    while( (ch=getchar()) < \'0\' || ch > \'9\');
        res = ch - 48;
    while( (ch = getchar() ) >= \'0\' && ch <= \'9\')
        res = ch - 48 + res * 10;
}
int DFS()
{
    int mi;
    for (register int i = 1; i <= n; i++)
    {
        for (register int j = t+1; j <= n; j++)
        {
            if (color[j] == color[t])
            {
                mi = 233333333;
                for (int e = t; e <= j; e++)
                {
                    if (house[e] < mi)
                    {
                        mi = house[e];
                    }
                }
                if (mi <= p) ans++;
            }
        }
        t++;
    }
    return ans;
}
int main()
{
    read(n);read(k);read(p);
    for (register int i = 1; i <= n; i++)
    {
        read(color[i]);
        read(house[i]);
    }
    cout << DFS();
    return 0;
}

#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int Count, sum, n, k, p, tmp;
int cnt[100];
int col[200010];
bool flag[200010];
int main()
{
    scanf("%d%d%d", &n, &k, &p);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &col[i]);
        scanf("%d", &tmp);
        if(tmp > p) flag[i] = 1;
    }
    for(int i = 1; i <= n; i++)
    {
        if(flag[i] == 0)
        {
            memset(cnt, 0, sizeof(cnt));
            continue;
        }
        cnt[col[i]]++;
        Count += (cnt[col[i]] - 1);
    }
    memset(cnt, 0, sizeof(cnt));
    for(int i = 1; i <= n; i++)
    {
        cnt[col[i]]++;
        sum += (cnt[col[i]] - 1);
    }
    cout << sum-Count;
    return 0;
}

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
int n;
int Num[15], i[6][8], Ans[6][3];
bool DEL(int i[6][8]) //清除
{
    bool Flag = 0, f[6][8]={0};
    for (int a = 1; a <= 5; a++)
      for (int b = 1; b <= 7; b++)
        if (i[a][b])
        {
            if (a <= 3 && i[a][b] == i[a+1][b] && i[a+1][b] == i[a+2][b])
              f[a][b] = f[a+1][b] = f[a+2][b] = true;
            if (b <= 5 && i[a][b] == i[a][b+1] && i[a][b+1] == i[a][b+2])
              f[a][b] = f[a][b+1] = f[a][b+2] = true;
        }
    for (int a = 1; a <= 5; a++)
      for (int b = 1; b <= 7; b++)
        if (f[a][b])
        {
            i[a][b] = 0;
            Flag = true;
        }
    return Flag;
}
void Falldown(int i[6][8]) //下落
{
    for (int a = 1; a <= 5; a++) //填补
    {
        int t = 0;
        for (int b = 1; b <= 7; b++)
        {
            int T = i[a][b];
            i[a][b] = 0;
            if (T)
              i[a][++t] = T;
        }
    }
}
bool Check(int i[6][8]) //检验
{
    for (int a = 1; a <= 5; a++)
      for (int b = 1; b <= 7; b++)
        if (i[a][b])
        return false;
    return true;
}
void DFS(int T, int i[6][8]) //DFS
{
    if (T > n) //超步子 
    {
        if (Check(i))
        {
            for (int a = 1; a <= n; a++)
              if (Ans[a][2]) //判断左右
                printf("%d %d -1\n", Ans[a][0], Ans[a][1]-1);
              else
                printf("%d %d 1\n", Ans[a][0]-1, Ans[a][1]-1);
            exit(0);//结束 
        }
        return;
    }
    memset(Num, 0, sizeof(Num)); //存储某种颜色格子现存数
    for (int a = 1; a <= 5; a++)
      for (int b = 1; b <= 7; b++)
        if (i[a][b])
          Num[i[a][b]]++;
    for (int a = 1; a <= 10; a++)
      if (Num[a] == 1 || Num[a] == 2) //等于0时并不需要返回
        return;
    int f[6][8] = {0}; //临时数组
    for (int a = 1; a < 5; a++)
      for (int b = 1; b <= 7; b++)
        if (i[a][b] != i[a+1][b]) //颜色不一样才交换 
        {
            memcpy(f, i, sizeof(f));//复制值 
            Ans[T][0] = a;
            Ans[T][1] = b;
            Ans[T][2] =! i[a][b]; //若空，即是true，若有，即是false
            swap(f[a][b], f[a+1][b]); //交换
            Falldown(f);
            while (DEL(f)) //消除
              Falldown(f);
            DFS(T+1, f);
        }
}
int main()
{
    scanf("%d", &n);
    for (int a = 1; a <= 5; a++)
      for (int b = 1; ; b++)
      {
        scanf("%d", &i[a][b]); //i列j行 
        if (!i[a][b]) //存储颜色
          break;
      }
    DFS(1, i);
    printf("-1");//执行至此如果没有结束则输出-1 
    return 0;
}