codeforces 835C 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

Copy

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

Copy

3
0
3

input

Copy

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

Copy

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：给你一个范围为100*100的天空，其中有很多星星，输入：n行为星星的亮度值，m行为查询时间t时区间（x1，y1) （x2，y2）的矩形范围内的星星的亮度值总和

题解：用一个三维数组sum[i][i][k]来表示在坐标（i，j）内的亮度为k的星星的数量，相当于二维前缀和表示了到i，j这个坐标的星星总数，再根据二维前缀和的容斥公式即求得

最后是O（C）的查询，c为星星亮度的最大值

代码如下：

#include<bits/stdc++.h>
using namespace std;
int mp[105][105][11];
int sum[105][105][11];
int main(){
    int n,q,c;
    int x,y,s;
    int t,x1,y1,x2,y2;
    while(scanf("%d%d%d",&n,&q,&c) !=EOF){
        memset(mp,0,sizeof(mp));
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;i++){
            scanf("%d%d%d",&x,&y,&s);
            mp[x][y][s]++;
        }
        for(int k=0;k<=c;k++){
            for(int i=1;i<=100;i++){
                for(int j=1;j<=100;j++){
                    sum[i][j][k]=sum[i-1][j][k]+sum[i][j-1][k]-sum[i-1][j-1][k]+mp[i][j][k];
                }
            }
        }
        for(int i=0;i<q;i++){
            scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);
            int ans=0;
            int ans1;
            for(int k = 0; k <= c; k++) {  
                ans1 = (k+t)%(c+1);        //    计算亮度为k的星星在t秒后的亮度  
                int num = sum[x2][y2][k]-sum[x1-1][y2][k]-sum[x2][y1-1][k]+sum[x1-1][y1-1][k];  //计算 （x1,y1）到（x2，y2）矩阵内亮度为k的星星的个数  
                ans1 *= num;  
                ans += ans1;  
            } 
            cout<<ans<<endl; 
        }
    }

}