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 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.

写到一半断网了。。。很绝望啊

容易发现c++后取模

应该是每一个(x+t%c)%c

但是我试图把几个x合起来

然后直接(sumx+有几个*t)然后去%c

很明显是不对的

只可以把%拆到括号里而不能把括号里的取出来

所以我们要另开一维 t%c

这样就可以分开存了

还有注意注意矩形是f[t][x2][y2]-f[t][x1-1][y2]-f[t][x2][y1-1]+f[t][x1-1][y1-1]

#include<cstdio>
#include<cstdlib>
int f[15][105][105];
int main()
{
	int n,q,c;
	scanf("%d %d %d",&n,&q,&c);
	c++;
	int x,y,s;
	for(int i=1;i<=n;i++)
	{
		scanf("%d %d %d",&x,&y,&s);
		for(int t=0;t<c;t++)
		f[t][x][y]+=(s+t)%c;
	}
	for(int t=0;t<c;t++)
	for(int i=1;i<=100;i++)
	for(int j=1;j<=100;j++)
	f[t][i][j]+=f[t][i-1][j]+f[t][i][j-1]-f[t][i-1][j-1];
	int t,x1,y1,x2,y2;
	for(int i=1;i<=q;i++)
	{
		scanf("%d %d %d %d %d",&t,&x1,&y1,&x2,&y2);
		t%=c;
		printf("%d\n",f[t][x2][y2]-f[t][x1-1][y2]-f[t][x2][y1-1]+f[t][x1-1][y1-1]);
	}
	return 0;
}