【Codeforces 835 C. Star sky】+ dp

C. Star sky

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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.

题意 ： 有 n 颗星星，给出每颗星星的坐标和亮度，q 次查询，给出左上角 和右下角的坐标问第 t 秒矩形内所有星星的亮度和，初始亮度为 o 的星星，第 t 秒的亮度为 （o + t）% （c + 1)

思路 ： dp ~ dp[x][y][k] 表示 1 ~ x,1~y这个矩形内亮度为 k 的星星的个数，然后 dp[x][y][k] = dp[x - 1][y][k] + dp[x][y - 1][k] - dp[x - 1][y - 1]k,查询时 ： ans =（o + t）% （c + 1) * dp[x2][y2][o] - dp[x1 - 1][y2][o] - dp[x2][y1 - 1][o] + dp[x1 - 1][x2 - 1][o] (重复减去的部分)

AC代码：

#include<cstdio>
typedef long long LL;
LL dp[110][110][12];
int main()
{
int n,q,c,x,y,s;
scanf("%d %d %d",&n,&q,&c);
for(int i = 0; i < n; i++)
scanf("%d %d %d",&x,&y,&s),dp[x][y][s]++;
for(int k = 0; k <= c; k++)
for(int i = 1; i <= 100; i++)
for(int j = 1; j <= 100; j++)
dp[i][j][k] += dp[i - 1][j][k] + dp[i][j - 1][k] - dp[i - 1][j - 1][k];
while(q--){
int t,x1,y1,x2,y2;
scanf("%d %d %d %d %d",&t,&x1,&y1,&x2,&y2);
LL ans = 0;
for(int i = 0; i <= c; i++)
ans += (LL)(dp[x2][y2][i] - dp[x1 - 1][y2][i] - dp[x2][y1 - 1][i] + dp[x1 - 1][y1 - 1][i]) * ((i + t) % (c + 1));
printf("%lld\n",ans);
}
return 0;
}