Star sky

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 (x_i, y_i), a maximum brightness c, equal for all stars, and an initial brightness s_i (0 ≤ s_i ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness s_i. 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 t_i and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x_1i, y_1i) and the upper right — (x_2i, y_2i). 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 x_i, y_i, s_i (1 ≤ x_i, y_i ≤ 100, 0 ≤ s_i ≤ 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 t_i, x_1i, y_1i, x_2i, y_2i (0 ≤ t_i ≤ 10⁹, 1 ≤ x_1i < x_2i ≤ 100, 1 ≤ y_1i < y_2i ≤ 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.

【题意】

有一片100*100的星空，上面有n颗星星，每个星星有一个亮度，且在0~C范围内周期性变化，现在给出q个查询，每个查询给出时间和一个矩形，求在该时间时矩形内星星的亮度和。

一开始看起来很难，就没怎么想，然后看懂了题意，就容易多了，dp就可以了。

 tree[i][j][k] = tree[i][j][k] + tree[i - 1][j][k] + tree[i][j - 1][k] - tree[i - 1][j - 1][k];
表示在坐标为（i,j)星星亮度为k的个数。
查询直接ans=tree[xx][yy][t] + tree[x - 1][y - 1][t] - tree[x - 1][yy][t] - tree[xx][y - 1][t];
当然还要考虑经过的时间t.

#include <bits/stdc++.h>
#define N 105
using namespace std;
int tree[N][N][12];
int n, q, c;

void dp() {
    for (int i = 1; i <= 100; i++) {
        for (int j = 1; j <= 100; j++) {
            for (int k = 0; k <= c; k++) {
                tree[i][j][k] = tree[i][j][k] + tree[i - 1][j][k] + tree[i][j - 1][k] - tree[i - 1][j - 1][k];
            }
        }
    }
}

int add(int t, int x, int y, int xx, int yy) {
    return tree[xx][yy][t] + tree[x - 1][y - 1][t] - tree[x - 1][yy][t] - tree[xx][y - 1][t];
}

int main() {

    while (scanf("%d%d%d", &n, &q, &c)!=EOF) {
        memset(tree, 0, sizeof(tree));
        for (int i = 0; i < n; i++) {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            tree[x][y][z]++;
        }
        dp();
        for (int i = 0; i < q; i++) {
            int t, x, y, xx, yy;
            scanf("%d%d%d%d%d", &t, &x, &y, &xx, &yy);
            int ans = 0;
            for (int j = 0; j <= c; j++) {
                int s = (t + j) % (c + 1);
                ans += s * add(j, x, y, xx, yy);
            }
            printf("%d
", ans);
        }
    }

    return 0;
}