Codeforces Round #427 (Div. 2)C. Star sky(dp)

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.

dp 问题   ， 转移方程   f[i][j][k]+=f[i][j-1][k]+f[i][j][k-1]-f[i][j-1][k-1];

坑点： 同一个位置居然可以有多个星星.....

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<ctime>#include<cstdlib>#include<utility>using namespace std;int f[11][110][110];int main(){int n,q,c;scanf("%d%d%d",&n,&q,&c);c++;int x,y,s;int i,j,k;memset(f,0,sizeof f);for(i=1;i<=n;i++){scanf("%d%d%d",&x,&y,&s);for(j=0;j<c;j++)f[j][x][y]+=(s+j)%c; // +=，重复的星星}for(i=0;i<c;i++)for(j=1;j<=100;j++)for(k=1;k<=100;k++)f[i][j][k]+=f[i][j-1][k]+f[i][j][k-1]-f[i][j-1][k-1];int t,x1,y1,x2,y2;for(i=1;i<=q;i++){scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);t%=c;int ans=f[t][x2][y2]-f[t][x1-1][y2]-f[t][x2][y1-1]+f[t][x1-1][y1-1];printf("%d\n",ans);}return 0;}