Codeforces Round #427 (Div.2) C

C. Star sky

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

【解题报告】
题意，n个点，给出坐标和初始亮度，最开始是0时刻，现在问你t时刻矩阵x1 y1 x2 y2 中亮度和是多少，每过一秒每个点的亮度加1，如果超过c则变成0。
简单Dp

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 110int n,q,c;  int dp[N][N][N];  int main(){    scanf("%d%d%d",&n,&q,&c);    for(int i = 1; i <= n; i ++)     {        int x,y,s;        scanf("%d%d%d",&x,&y,&s);        for(int j=0;j<=c;++j)         {            dp[j][x][y]+=(s+j)%(c+1);        }    }    for(int i=1;i<=100;++i)     for(int j=1;j<=100;++j)     for(int k=0;k<=c;++k)     {        dp[k][i][j]+=dp[k][i-1][j]+dp[k][i][j-1]-dp[k][i-1][j-1];       }    for(int i=1;i<=q;++i)    {        int x1,x2,y1,y2,t;        scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);        t%=(c+1);        printf("%d\n",dp[t][x2][y2]-dp[t][x2][y1-1]-dp[t][x1-1][y2]+dp[t][x1-1][y1-1]);    }    return 0;}