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.
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.StringTokenizer;

public class C {
	public static void main(String [] arg) throws Exception, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer tokenizer = new StringTokenizer(br.readLine());
		
		int n = Integer.parseInt(tokenizer.nextToken());
		int numQueries = Integer.parseInt(tokenizer.nextToken());
		int maxLight = Integer.parseInt(tokenizer.nextToken());
		int x,y,light;
		int [][][] T = new int[maxLight+1][105][105];
		for(int i = 0; i < n; i++) {
			tokenizer = new StringTokenizer(br.readLine());
			x = Integer.parseInt(tokenizer.nextToken());
			y = Integer.parseInt(tokenizer.nextToken());
			light = Integer.parseInt(tokenizer.nextToken());
			T[light][y][x]++;
		}

		
		for(int step = 0; step <= maxLight; step++) {
			for(int i = 1; i<105; i++) {
				for(int j = 1; j < 105; j++) {
					T[step][i][j] += T[step][i-1][j] + T[step][i][j-1] - T[step][i-1][j-1];
				}
			}
		}

		int x1,y1,x2,y2,t;
		BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
		for(int i = 0; i < numQueries; i++) {
			tokenizer = new StringTokenizer(br.readLine());
			t = Integer.parseInt(tokenizer.nextToken());
			x1 = Integer.parseInt(tokenizer.nextToken());
			y1 = Integer.parseInt(tokenizer.nextToken());
			x2 = Integer.parseInt(tokenizer.nextToken());
			y2 = Integer.parseInt(tokenizer.nextToken());

			bw.write(count(x1, y1, x2, y2, t, maxLight, T) + "");
			bw.newLine();
		}
		bw.close();
	}
	static int count(int x1, int y1, int x2, int y2, int step, int maxLight, int [][][] T) {
		int res = 0;
		for(int i = 0; i <= maxLight; i++) {
			int brightness = (i + step) % (maxLight+1);
			int numStars =  T[i][y2][x2] - T[i][y2][x1-1] - T[i][y1-1][x2] + T[i][y1-1][x1-1];
			res +=  brightness*numStars;
		}
		return res;
	}

}