Problem: Nuske has a grid with N rows and M columns of squares. The rows are numbered 1 through N from top to bottom, and the columns are numbered 1 through M from left to right. Each square in the grid is painted in either blue or white. If S_{i,j} is 1, the square at the i-th row and j-th column is blue; if S_{i,j} is 0, the square is white. For every pair of two blue square a and b, there is at most one path that starts from a, repeatedly proceeds to an adjacent (side by side) blue square and finally reaches b, without traversing the same square more than once.

Phantom Thnook, Nuske's eternal rival, gives Q queries to Nuske. The i-th query consists of four integers x_{i,1}, y_{i,1}, x_{i,2} and y_{i,2} and asks him the following: when the rectangular region of the grid bounded by (and including) the x_{i,1}-th row, x_{i,2}-th row, y_{i,1}-th column and y_{i,2}-th column is cut out, how many connected components consisting of blue squares there are in the region?

Process all the queries.

Constraints

* 1 ≤ N,M ≤ 2000
* 1 ≤ Q ≤ 200000
* S_{i,j} is either 0 or 1.
* S_{i,j} satisfies the condition explained in the statement.
* 1 ≤ x_{i,1} ≤ x_{i,2} ≤ N(1 ≤ i ≤ Q)
* 1 ≤ y_{i,1} ≤ y_{i,2} ≤ M(1 ≤ i ≤ Q)

Input

The input is given from Standard Input in the following format:


N M Q
S_{1,1}..S_{1,M}
:
S_{N,1}..S_{N,M}
x_{1,1} y_{i,1} x_{i,2} y_{i,2}
:
x_{Q,1} y_{Q,1} x_{Q,2} y_{Q,2}


Output

For each query, print the number of the connected components consisting of blue squares in the region.

Examples

Input

3 4 4
1101
0110
1101
1 1 3 4
1 1 3 1
2 2 3 4
1 2 2 4


Output

3
2
2
2


Input

5 5 6
11010
01110
10101
11101
01010
1 1 5 5
1 2 4 5
2 3 3 4
3 3 3 3
3 1 3 5
1 1 3 4


Output

3
2
1
1
3
2

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,m,t1,ans,a[2010][2010],b[2010][2010],c[2010][2010];
char s[2010];
int geta(int l1,int r1,int l2,int r2){
	return a[l2][r2]-a[l2][r1-1]-a[l1-1][r2]+a[l1-1][r1-1];
}
int getb(int l1,int r1,int l2,int r2){
	return b[l2][r2]-b[l2][r1-1]-b[l1-1][r2]+b[l1-1][r1-1];
}
int getc(int l1,int r1,int l2,int r2){
	return c[l2][r2]-c[l2][r1-1]-c[l1-1][r2]+c[l1-1][r1-1];
}
int main(){
	int i,j,l1,r1,l2,r2;
	scanf("%d%d%d",&n,&m,&t1);
	for(i=1;i<=n;i++){
		scanf("%s",s+1);
		for(j=1;j<=m;j++)a[i][j]=s[j]-'0';
	}
	for(i=1;i<=n;i++)
	 for(j=2;j<=m;j++)if(a[i][j]&&a[i][j-1])b[i][j]=1;
	for(i=2;i<=n;i++)
	 for(j=1;j<=m;j++)if(a[i][j]&&a[i-1][j])c[i][j]=1;
	for(i=1;i<=n;i++)
	 for(j=1;j<=m;j++)
	  b[i][j]=b[i][j]+b[i-1][j]+b[i][j-1]-b[i-1][j-1],
	  c[i][j]=c[i][j]+c[i-1][j]+c[i][j-1]-c[i-1][j-1],
	  a[i][j]=a[i][j]+a[i-1][j]+a[i][j-1]-a[i-1][j-1];
	while(t1--){
		scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
		ans=geta(l1,r1,l2,r2);
		ans-=getb(l1,r1+1,l2,r2);
		ans-=getc(l1+1,r1,l2,r2);
		printf("%d\n",ans);
	}
}
/*
是棵树
那么sz=总-边数
边数就是相邻两个 
*/