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.

题意：给出二维面上星星的亮度，每次询问t秒后，某区域内星星亮度总和，其中亮度变化s=(s+t)%(c+1)

思路：一个点上可能有多个点，定义一个数组d[c][x][y]，表示（0,0）->（x,y）区域上亮度为c的总个数，状态转移方程为：dp[k][i][j] += dp[k][i - 1][j] + dp[k][i][j - 1] - dp[k][i - 1][j - 1]。利用这些前缀和相减，就能使每次询问得出结果复杂程度降到O（1）复杂度

上代码

#include<iostream>#include<cmath>#include<cstring>#include<cstdio>#include<string>#include<queue>#include<stack>#include<vector>#include<map>#include<algorithm>using namespace std;#define ll long long#define inf 0x3f3f3f3f#define ls l,m,rt<<1#define rs m+1,r,rt<<1|1#define maxn 105int dp[12][maxn][maxn];int main(){//freopen("Text.txt","r",stdin);int n, q, c,x,y,s,x1,y1,t;memset(dp, 0, sizeof(dp));scanf("%d%d%d", &n, &q, &c);while (n--){scanf("%d%d%d", &x, &y, &s);dp[s][x][y]++;}for (int i = 1; i <= 100; i++){for (int j = 1; j <= 100; j++){for (int k = 0; k <= c; k++){dp[k][i][j] += dp[k][i - 1][j] + dp[k][i][j - 1] - dp[k][i - 1][j - 1];//是指（0,0）到（x,y）区域内亮度为k的总个数    //printf("%d ", dp[k][i][j]);    }//printf("\n");}//printf("\n");}while (q--){int sum[12],cnt=0;scanf("%d%d%d%d%d", &t, &x, &y, &x1, &y1);for (int i = 0; i <= c; i++){sum[i] = dp[i][x1][y1] + dp[i][x - 1][y - 1] - dp[i][x1][y - 1] - dp[i][x - 1][y1];//画图好理解些，因为边缘点要包括cnt += ((i + t)%(c+1))*sum[i];}printf("%d\n", cnt);}return 0;}