Codeforces Round #427 (Div. 2) C. 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

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.

这道题暴力去写肯定会超时的，所以要缩短去查找所需要的时间。

dp[i][j][k]表示从（0，0）到（i，j）的矩形中所存在的亮度为k的星星的数量。

由容斥原理（初中平面几何）可得：dp[i][j][k]+=dp[i-1][j][k]+dp[i][j-1][k]-dp[i-1][j-1][k];

这样可以大幅度减少查找的工作量，得到结果。

AC代码：

//_bread#include<iostream>#include<algorithm>#include<cstring>#include<string>using namespace std;int dp[111][111][15];int main(){int n,q,c;int x,y,s;memset(dp,0,sizeof(dp));cin>>n>>q>>c;for(int i=0;i<n;i++){cin>>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<=c;k++)dp[i][j][k]+=dp[i-1][j][k]+dp[i][j-1][k]-dp[i-1][j-1][k];}}int x1,y1,x2,y2,ss;while(q--){cin>>ss>>x1>>y1>>x2>>y2;int res=0;for(int i=0;i<=c;i++){int k=(i+ss)%(c+1);res+=k*(dp[x2][y2][i]-dp[x2][y1-1][i]-dp[x1-1][y2][i]+dp[x1-1][y1-1][i]);}cout<<res<<endl;}return 0;}