#include<bits/stdc++.h>
using namespace std;
const int N = 1005, M = 55;
const int INF = 1e8;
#define x first;
#define y second;
//数的棵树，绿化图大小，藏宝图大小
int n, L, S;
pair<int, int> tree[N];
int B[M][M];

int main() {
    cin >> n >> L >> S;
    for (int i = 0; i < n; i++) {
        int x1, y1;
        cin >> x1 >> y1;
        tree[i] = {x1, y1};
    }
    int Bsize = 0; //B中树的数量
    for (int i = S; i >= 0; i--) {
        for (int j = 0; j <= S; j++) {
            cin >> B[i][j];
            Bsize += B[i][j];
        }
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
        //遍历绿化图中的每一颗树，以这颗树作为原点划分一个S*S大小的区域A
        int curX = tree[i].x;
        int curY = tree[i].y;
        //首先判断当前当前区域是否会越界
        if (curX + S > L || curY + S > L)continue;
        int cnt = 0; //当前A区域树的个数
        //遍历绿化图所有的树，统计当前A区域中树的数量以及是否与对应的B匹配
        for (int j = 0; j < n; j++) {
            int curX1 = tree[j].x;
            int curY1 = tree[j].y;
            //首先判断当前树是否在A区域中
            if (curX1 >= curX && curX + S >= curX1 && curY1 >= curY && curY + S >= curY1) {
                //如果对应的B中没有树,置为负无穷
                if (!B[curX1 - curX][curY1 - curY]) cnt = -INF;
                else cnt++;
            }
        }
        if (cnt == Bsize) ans++;
    }
    cout << ans << endl;
}