Codeforces 835 C 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 (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.

题目大意：

在一个笛卡尔坐标系中，有n颗星星，然后每颗星星都有自己的初始亮度，然后整个星系中都有最亮的亮度，星星的亮度会随着时间而增加，但是到达最亮值的下一个单位的时间就会变成0，然后继续。。。。题目让我们求在一个矩阵里的所有星星T秒后的亮度总和

解题思路：

先把图存下来，然后能解出其他秒数时候的图，之后求矩阵和，，注意，同一个点可能有两颗星星

AC代码：

#include<stdio.h>

int a[201][201][20];

int main(){
    int n, m, k;
    scanf("%d%d%d", &n, &m, &k);
    for(int i = 0; i < n; i++){
        int x, y, z;
        scanf("%d%d%d", &x, &y, &z);
        for(int j = 0; j <= k; j++){
            a[x][y][j] += (z + j) % (k + 1);
        }
    }
    for(int i = 1; i <= 100; i ++){
        for(int j = 1; j <= 100; j++){
            for(int l = 0; l <= k; l++){
                a[i][j][l] += a[i - 1][j][l] +  a[i][j - 1][l] - a[i - 1][j - 1][l];
            }
        }
    }
    int t, x1, y1, x2, y2;
    while(m--){
        scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);
        t %= (k + 1);
        printf("%d\n", a[x2][y2][t] - a[x2][y1 - 1][t] - a[x1 - 1][y2][t] + a[x1 - 1][y1 - 1][t]);
    }
}