codeforces 835-C. Star sky（dp+前缀和）

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.

题目大意：

在x*y的格子中（最大规模为100*100），有n个星星，每个星星都有一个发光亮度。并且这种发光亮度随着时间增长，但是不会超过给定值C，如果超过了，就会重新从0开始增长。现在给出q个询问，每个询问给出一个矩形和时间，问在矩形内的星星发光总亮度为多少？

解题思路：

因为每个星星带有亮度的属性，因此在压缩点的时候需要多开一维。

用a[i][j][k] 表示点（i，j）发光亮度为k的星星个数。

接着因为有q次询问，考虑到会超时，需要用到前缀和。

dp[k][i][j] 表示 第 i 行前 j 个点中发光亮度为k 的星星的个数

#include<iostream>
#include<cstring>
using namespace std;
int a[105][105][15],dp[15][105][105];    // 第i行前j个点中s为k的点的个数
int main()
{
int n,q,c,x,y,s;
while(cin>>n>>q>>c)
{
memset(a,0,sizeof(a));
memset(dp,0,sizeof(dp));
for(int i=0;i<n;i++)
{
cin>>x>>y>>s;
a[x][y][s]++;
}
for(int k=0;k<=10;k++)
for(int i=1;i<=100;i++)
for(int j=1;j<=100;j++)
dp[k][i][j]=dp[k][i][j-1]+a[i][j][k];
int t,x1,x2,y1,y2;
for(int i=0;i<q;i++)
{
cin>>t>>x1>>y1>>x2>>y2;
int ans=0;
for(int k=0;k<=10;k++)
for(int j=x1;j<=x2;j++)
ans+=(dp[k][j][y2]-dp[k][j][y1-1])*((k+t)%(c+1));
cout<<ans<<endl;
}
}
return 0;
}