CodeForces

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_iand 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.

Example

Input

2 3 31 1 13 2 02 1 1 2 20 2 1 4 55 1 1 5 5

Output

303

Input

3 4 51 1 22 3 03 3 10 1 1 100 1001 2 2 4 42 2 1 4 71 50 50 51 51

Output

3350

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.

注意星星到达最大亮度之后就会变回0；

开三维数组第一个变量代表时间。

#include<iostream>  #include<cstdio>  #include<cstring>  #include<algorithm>#include<cmath>  using namespace std;int ma[12][110][110];int  n ,q ,c;int main(){scanf("%d%d%d", &n, &q, &c);int x, y, s;for(int i = 0; i < n; i++){scanf("%d%d%d", &x,&y ,&s);for(int k = 0; k <= 10; k++){     //初始化11个时间 ma[k][x][y] += (s + k) % ( c + 1);   //应该是+=  }}for(int k = 0; k <= 10; k++)for(int i = 1; i <= 100; i++)for(int a = 1; a <= 100; a++){ma[k][i][a] += ma[k][i][a-1] + ma[k][i-1][a] - ma[k][i-1][a-1];      //写错加等于自己 } int x1, y1, x2, y2, ti;while(q--){scanf("%d%d%d%d%d", &ti, &x1, &y1, &x2, &y2);ti %= (c + 1);int te = ma[ti][x2][y2] - ma[ti][x1 -1][y2] -ma[ti][x2][y1-1] + ma[ti][x1 -1][y1 - 1];printf("%d\n", te);}return 0;}