Codeforces Round #427 (Div. 2)-C. Star sky

题目：                                              C. Star sky

The Cartesian coordinate system is set in the sky. There you can seen stars, the i-th has coordinates (x_i,y_i), a maximum brightnessc, equal for all stars, and an initial brightnesss_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, and0, otherwise.

You want to look at the sky q times. In the i-th time you will look at the momentt_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. Thei-th from these lines contains three integersx_i,y_i,s_i (1 ≤ x_i, y_i ≤ 100,0 ≤ s_i ≤ c ≤ 10) — the coordinates ofi-th star and its initial brightness.

The next q lines contain the views description. Thei-th from these lines contains five integerst_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 thei-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 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 moment2 its brightness is 3, so the answer is3.

At the second view, you can see only the second star. At moment0 its brightness is 0, so the answer is0.

At the third view, you can see both stars. At moment5 brightness of the first is 2, and brightness of the second is 1, so the answer is3.

题意：给出n个星星的坐标(x_i,y_i)和初始亮度s_i。星星在t秒时的亮度为(s_i+t)%(c+1)。求在t秒时以(x1,y1)为左下角、(x2,y2)为右上角的矩形内星星的总亮度。

思路：比赛时用暴力做，总是在第八组数据超时，郁闷。早上醒来就去搜索大神博客看题解，才明白我有多笨。。。用一个三维数组a[i][j][k]记录以(i,j)为右上角的矩形内星星亮度为k的数量，然后答案就是a[x2][y2][k]-a[x1-1][y2][k]-a[x2][y1-1][k]+a[x1-1][y1-1][k]。一切玄妙尽在代码中......

Code：

#include<bits/stdc++.h>using namespace std;int a[105][105][15];int get(int x1,int y1,int x2,int y2,int k){        return a[x2][y2][k]-a[x1-1][y2][k]-a[x2][y1-1][k]+a[x1-1][y1-1][k];}int main(){        int n,q,c,x,y,s,i,j,k,t,x1,y1,x2,y2,sum;        while(~scanf("%d%d%d",&n,&q,&c)){                memset(a,0,sizeof(a));                for(i=0;i<n;i++){                        scanf("%d%d%d",&x,&y,&s);                        a[x][y][s]++;                }                c++;                for(i=1;i<101;i++){   //求以(i,j)为右上角的矩形内星星亮度为k的数量                        for(j=1;j<101;j++){                                for(k=0;k<c;k++) a[i][j][k]+=a[i-1][j][k]+a[i][j-1][k]-a[i-1][j-1][k];                        }                }                while(q--){                        scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);                        sum=0;                        for(i=0;i<c;i++) sum+=get(x1,y1,x2,y2,i)*((i+t)%c);                        printf("%d\n",sum);                }        }    return 0;}