CF-835C

C. Star sky

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.

You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.

A star lies in a rectangle if it lies on its border or lies strictly inside it.

Input

The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.

The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si (1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.

The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.

Output

For each view print the total brightness of the viewed stars.

Examples

input

2 3 3
1 1 1
3 2 0
2 1 1 2 2
0 2 1 4 5
5 1 1 5 5

output

3
0
3

input

3 4 5
1 1 2
2 3 0
3 3 1
0 1 1 100 100
1 2 2 4 4
2 2 1 4 7
1 50 50 51 51

output

3
3
5
0

Note

Let's consider the first example.

At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.

At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.

At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.

求出所有点的前缀和，之后对于每个t，计算sum[t][x2][y2]-sum[t][x1-1][y2]-sum[t][x2][y1-1]+sum[t][x1-1][y1-1]并输出。

AC代码：

#include<bits/stdc++.h>
using namespace std;

const int MAXN=110;

int sum[15][MAXN][MAXN];

int main(){
    ios::sync_with_stdio(false);
    int n,q,c;
    cin>>n>>q>>c;
    for(int i=0;i<n;i++){
        int x,y,s;
        cin>>x>>y>>s;
        for(int t=0;t<=c;t++){
            sum[t][x][y]+=(s+t)%(c+1);
        }
    }
    for(int i=0;i<=c;i++){
        for(int x=1;x<MAXN;x++){
            for(int y=1;y<MAXN;y++){
                sum[i][x][y]+=sum[i][x-1][y]+sum[i][x][y-1]-sum[i][x-1][y-1];
            }
        }
    }
    for(int i=0;i<q;i++){
        int t,x1,y1,x2,y2;
        cin>>t>>x1>>y1>>x2>>y2;
        t%=(c+1);
        cout<<sum[t][x2][y2]-sum[t][x1-1][y2]-sum[t][x2][y1-1]+sum[t][x1-1][y1-1]<<endl;;
    }
    return 0;
}