Codeforces Round #427 (Div. 2)

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 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.

题意：有n颗星星，每颗星星有各自的初始位置（xi,yi）以及初始亮度 si ，所有星星的最大亮度均为 c ，星星亮度每秒+1，当亮度大于 c 时变为 0 ，求第 t 秒时一个矩形区域内所有的星星的亮度总和。

因为星星的亮度变化是周期性的，所以任一区域内的星星亮度总和也是周期性的，周期长度为 c+1 ，（x1,y1）到（x2,y2）矩形区域内星星的亮度总和可以由下图所示的大矩形减小矩形得到

                     

所以我们只要把0-c秒内所有的（0,0）到（x,y）内所包含的所有星星的亮度总和求出，输出答案的时候加加减减一下就行了。

刚开始以为一个坐标只有一个星星一直错

#include<iostream>#include<string.h>#include<stdio.h>using namespace std;int n,q,s,sum[15][111][111];//sum[t][x][y]代表在t秒时(0,0)到(x,y)矩形内所有星星的亮度总和int main(){    while(~scanf("%d%d%d",&n,&q,&s))    {        memset(sum,0,sizeof sum);        for(int i=0;i<n;i++)        {            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            for(int j=0;j<=10;j++)                sum[j][a][b] += (c + j) % (s+1);//一个位置可以出现多个星星        }        for(int i=0;i<=10;i++)            for(int j=1;j<=100;j++)                for(int k=1;k<=100;k++)                    sum[i][j][k] += sum[i][j-1][k] + sum[i][j][k-1] - sum[i][j-1][k-1];        while(q--)        {            int t,x1,x2,y1,y2;            scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);            t %= (s+1);            int ans = sum[t][x2][y2] + sum[t][x1-1][y1-1] - sum[t][x1-1][y2] - sum[t][x2][y1-1];            printf("%d\n",ans);        }    }    return 0;}