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There are many rook on a chessboard, a rook can attack the row and column it belongs, including its own place.

There are also many queries, each query gives a rectangle on the chess board, and asks whether every grid in the rectangle will be attacked by any rook?

InputThe first line of the input is a integer

TT, meaning that there are

TT test cases.

Every test cases begin with four integers

n,m,K,Qn,m,K,Q.

KK is the number of Rook,

QQ is the number of queries.

Then

KK lines follow, each contain two integers

x,yx,y describing the coordinate of Rook.

Then

QQ lines follow, each contain four integers

x1,y1,x2,y2x1,y1,x2,y2 describing the left-down and right-up coordinates of query.

1≤n,m,K,Q≤100,0001≤n,m,K,Q≤100,000.

1≤x≤n,1≤y≤m1≤x≤n,1≤y≤m.

1≤x1≤x2≤n,1≤y1≤y2≤m1≤x1≤x2≤n,1≤y1≤y2≤m.

OutputFor every query output "Yes" or "No" as mentioned above.

Sample Input

22 2 1 21 11 1 1 22 1 2 22 2 2 11 11 22 1 2 2

Sample Output

YesNoYes

题目大意：棋盘上的一个车可以吃掉同一行同一列的棋子，告诉你车的坐标和一块区域，看看在这块区域中的棋子是不是都会被吃掉

前缀和？

树状数组？？

代码如下

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;const int maxn=100000+10;int row[maxn],col[maxn];int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n,m,k,q;        memset(row,0,sizeof(row));        memset(col,0,sizeof(col));        scanf("%d%d%d%d",&n,&m,&k,&q);        while(k--)        {            int a,b;            scanf("%d%d",&a,&b);            row[a]=1,col[b]=1;  //赋值为1  有可能出现相同的不能累加        }        for(int i=1;i<=n;i++)        {            row[i]+=row[i-1];        }        for(int i=1;i<=m;i++)        {            col[i]+=col[i-1];        }        while(q--)        {            int x1,y1,x2,y2,res1,res2;            scanf("%d%d%d%d",&x1,&y1,&x2,&y2);            res1=row[x2]-row[x1-1];            res2=col[y2]-col[y1-1];            if(res1==(x2-x1+1)||res2==(y2-y1+1))                printf("Yes\n");            else                printf("No\n");        }    }    return 0;}

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