// lanqiao 2109 统计子矩阵(二维前缀和)
#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N = 5e2 + 10;
int n, m, k;
int a[N][N], sum[N][N];
// 暴力解法，枚举子矩阵，二维前缀和求矩阵和，时间复杂度 O(n^4)
signed main(){
    cin >> n >> m >> k;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            cin >> a[i][j];
            sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + a[i][j];
        }
    }
    int ans = 0, res;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            for(int i1 = i; i1 <= n; i1++){
                for(int j1 = j; j1 <= m; j1++){
                    res  = sum[i1][j1] - sum[i-1][j1] - sum[i1][j-1] + sum[i-1][j-1];
                    if(res <= k) ans++;
                }
            }
        }
    }
    cout << ans << endl;
    return 0;
}
/* samples -> 19
3 4 10
1 2 3 4
5 6 7 8
9 10 11 12
*/