Codeforces 835C-Star sky

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.

题意：给出 n个星星的坐标和它们的亮度si，星星最大亮度为c，t秒后星星的亮度为（si+t）%（c+1），q次询问，每次询问一个区间内t秒后星星的亮度和

解题思路：开一个三维数组，记录每个前缀亮度为si的星星的个数即可

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <map>  #include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  #include <functional>using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;int a[105][105][15];int n, q, c;int x[100005], y[100005], s[100005];int main(){while (~scanf("%d%d%d", &n, &q, &c)){memset(a, 0,sizeof a);for (int i = 1; i<=n; i++){scanf("%d%d%d", &x[i], &y[i], &s[i]);a[x[i]][y[i]][s[i]]++;}for (int i = 0; i <= 10; i++)for (int j = 1; j<102; j++)for (int k = 1; k<102; k++)a[j][k][i] += (a[j - 1][k][i] + a[j][k - 1][i] - a[j - 1][k - 1][i]);while (q--){int t, x1, y1, x2, y2;scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2);int ans = 0;for (int i = 0; i <= 10; i++){int k = a[x2][y2][i] - a[x1 - 1][y2][i] - a[x2][y1 - 1][i] + a[x1 - 1][y1 - 1][i];int kk = (t + i) % (c + 1);ans += (k*kk);}printf("%d\n", ans);}}return 0;}