codeforces Star sky（二维前缀和）

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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：给定n个点，点的坐标为(x,y)，每个点上有一个亮度为s的星星，星星的亮度随时间的增加一秒而增加1，但是当星星的亮度达到c+1时置为0，然后有随时间的增加而增加。然后输入q次查询，每次查询的输入为：时间t，左下角的坐标(x1,y1)，右上角的坐标(x2,y2)，求问t时刻，在这个矩形内，所有星星的总亮度为多少？

题解：显然是一个二维前缀和，刚开始看错题意wa了一发。。。用一个三维的dp数组，dp[i][j][k]表示(0,0)到(i,j)的矩形内亮度为k的星星的个数，c小于等于10，所以无论何时星星的亮度不会超过10。t时刻，初始亮度为k的星星的亮度为(k+t)%(c+1)。这里注意二维前缀和的写法，最好画个图表示，不然很容易忽略掉某个区域。

#include<stdio.h>#include<iostream>#include<string>#include<string.h>#include<algorithm>#include<set>#include<stdlib.h>#include<queue>#include<vector>#include<math.h>using namespace std;typedef long long ll;int n , q , c;int dp[105][105][15];int main(){    int t , x1 , y1 , x2 , y2;    while(~scanf("%d%d%d" , &n , &q , &c)){        memset(dp , 0 , sizeof(dp));        for(int i = 0 ; i < n ; i ++){            int x , y , s ;            scanf("%d %d %d" , &x , &y , &s);            dp[x][y][s]++;        }        for(int i = 1 ; i <= 100 ; i ++){            for(int j = 1 ; j <= 100 ; j ++){                for(int k = 0 ; k < 15 ; k ++){                    dp[i][j][k] += dp[i-1][j][k] + dp[i][j-1][k] - dp[i-1][j-1][k];                }            }        }        while(q--){            scanf("%d%d%d%d%d" , &t , &x1 , &y1 , &x2 , &y2);            int ans = 0 ;            for(int k = 0 ; k < 15 ; k ++){                int pre = dp[x2][y2][k] - dp[x2][y1-1][k] - dp[x1-1][y2][k] + dp[x1-1][y1-1][k];                ans += pre*((k+t)%(c+1));            }            cout<<ans<<endl;        }    }    return 0;}