C. Star sky Codeforces

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 (x_i,y_i), a maximum brightnessc, equal for all stars, and an initial brightnesss_i (0 ≤ s_i ≤ c).

Over time the stars twinkle. At moment 0 the i-th star has brightness s_i. Let at momentt some star has brightness x. Then at moment (t + 1) this star will have brightnessx + 1, if x + 1 ≤ c, and0, otherwise.

You want to look at the sky q times. In thei-th time you will look at the moment t_i and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x_1i,y_1i) and the upper right — (x_2i,y_2i). 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. Thei-th from these lines contains three integersx_i,y_i,s_i (1 ≤ x_i, y_i ≤ 100,0 ≤ s_i ≤ c ≤ 10) — the coordinates ofi-th star and its initial brightness.

The next q lines contain the views description. Thei-th from these lines contains five integerst_i,x_1i,y_1i,x_2i,y_2i (0 ≤ t_i ≤ 10⁹,1 ≤ x_1i < x_2i ≤ 100,1 ≤ y_1i < y_2i ≤ 100) — the moment of thei-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 is2, and brightness of the second is 1, so the answer is 3.

题意：在坐标中给了你很多带有亮度的星星。这些星星随着时间做周期性变化。然后给出q个询问，每一个询问包含了查询区域的位置与当前时间，让你求出这段区域的总亮度是什么。

思路：

预处理到每一个从原点出发到每一个点所围的矩形内星星的个数。

dp【x】【y】【z】就是从1,1点出去到x，y所形成的矩阵中（包括边界）的亮度为z的星星的个数

所以dp【X】【Y】【Z】=dp【X】【Y】【Z】+dp【X-1】【Y】【C】+dp【X】【Y-1】【C】-DP【X-1】【Y-1】【C】

#include<bits/stdc++.h>using namespace std;int dp[120][120][11];int n,q,c;void init(){    for(int i=1; i<=100; i++)        for(int j=1; j<=100; j++)            for(int c=0; c<=10; c++)                dp[i][j][c]+=dp[i-1][j][c]+dp[i][j-1][c]-dp[i-1][j-1][c];}int main(){    while(scanf("%d%d%d",&n,&q,&c)!=EOF)    {        memset(dp,0,sizeof(dp));        for(int i=0; i<n; i++)        {            int x,y,b;            scanf("%d%d%d",&x,&y,&b);            dp[x][y][b]++;        }        init();        for(int j=0; j<q; j++)        {            int x,y,x1,y1,t;            int ans=0;            int sum=0;            scanf("%d%d%d%d%d",&t,&x,&y,&x1,&y1);            for(int i=0; i<=c; i++)            {                ans=dp[x1][y1][i]+dp[x-1][y-1][i]-dp[x-1][y1][i]-dp[x1][y-1][i];                ans*=(t+i)%(c+1);                sum+=ans;            }            printf("%d\n",sum);        }    }}