Codeforces835C 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.

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题目的意思是有一片100*100的星空，上面有n颗星星，每个星星有一个亮度，且在

0~C范围内周期性变化，现在给出q个查询，每个查询给出时间和一个矩形，求在该

时间时矩形内星星的亮度和。

思路：先预处理处每个区域每种亮度的星星有多少颗，可以直接搞波前缀和，然后

枚举各个亮度在查询区域内的星星个数，计算总和

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;#define mod 10000007#define mem(a,b) memset(a,b,sizeof a)int main(){int n,q,c,x,y,x1,x2,y1,y2,c1,t;int mp[105][105][15];while(~scanf("%d%d%d",&n,&q,&c)){    mem(mp,0);    for(int i=0;i<n;i++)    {        scanf("%d%d%d",&x,&y,&c1);        mp[x][y][c1]++;    }    for(int i=0;i<=10;i++)    {        for(int j=1;j<102;j++)        {            for(int k=1;k<102;k++)            {                mp[j][k][i]+=(mp[j-1][k][i]+mp[j][k-1][i]-mp[j-1][k-1][i]);            }        }    }    while(q--)    {        scanf("%d%d%d%d%d",&t,&x1,&y1,&x2,&y2);        int ans=0;        for(int i=0;i<=10;i++)        {            int cnt=mp[x2][y2][i]-mp[x1-1][y2][i]-mp[x2][y1-1][i]+mp[x1-1][y1-1][i];            int xx=(t+i)%(c+1);            ans+=(cnt*xx);        }        printf("%d\n",ans);    }}    return 0;}