cf 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.

题意：给你n个星星的坐标及亮度，q次查询，问在给定的矩形范围内星星亮度的总和是多少？c为最大亮度；

题解：刚开始的时候我以为星星的亮度变化是 0 ~ n ~ 0，其实是0 ~ n再从0 ~ n，然后一直错，这道题可以先与处理一下0 ~ c 每个亮度的星星有多少，然后求矩形的内部星星的亮度的总和和树状数组是相似的，t秒之后的星星的亮度为（t+si）%（c+1）

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; typedef long long ll; int dp[1000+10][1000+10][11]; main() {     int n,m,c;     while(cin>>n>>m>>c){         int x,y,v;         memset(dp,0,sizeof(dp));         for(int i=0;i<n;i++){             cin>>x>>y>>v;             dp[x][y][v]++;         }         for(int i=1;i<=100;i++){             for(int j=1;j<=100;j++){                 for(int k=0;k<=c;k++){                     dp[i][j][k]+=dp[i-1][j][k] + dp[i][j-1][k] -dp[i-1][j-1][k];                 }             }         }         while(m--){             int t,x1,y1,x2,y2;             cin>>t>>x1>>y1>>x2>>y2;             ll ans = 0;             for(int i=0;i<=c;i++){                 int pos = (i + t) %(c + 1);                 ans = ans + pos * (dp[x2][y2][i] - dp[x1-1][y2][i] - dp[x2][y1-1][i] + dp[x1-1][y1-1][i]);             }             cout<<ans<<endl;         }     } }