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.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class C {
	
	static int sum(int x1, int y1, int x2, int y2, int[][] sum){
		int s = sum[x2][y2];
		if(x1 - 1 >= 0 && y1 - 1 >= 0) s += sum[x1 - 1][y1 - 1];
		if(y1 - 1 >=0) s -= sum[x2][y1 - 1];
		if(x1 - 1 >=0) s -= sum[x1 - 1][y2];
		return s;
	}
	
	public static void main(String[] args) throws IOException {
		File inputFile = new File("entrada");
		if (inputFile.exists())
			System.setIn(new FileInputStream(inputFile));
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		StringBuilder out = new StringBuilder();
		String line;
		while ((line = in.readLine()) != null) {
			Integer[] nqc = readInts(line);
			int n = nqc[0], q = nqc[1],c = nqc[2] + 1;
			Integer[][] stars = new Integer[n][];
			int[][][] dp = new int[11][101][101];
			for(int i = 0; i < n; i++) {
				stars[i] = readInts(in.readLine());
				for(int j = 0, s = stars[i][2]; j <= 10; j++, s = (s + 1) % c) 
					dp[j][stars[i][0]][stars[i][1]] += s;
			}
			for(int d = 0; d <= 10; d++)
				for(int i = 0; i <= 100; i++)
					for(int j = 0; j <= 100; j++){
						if(i - 1 >= 0) dp[d][i][j] += dp[d][i - 1][j];
						if(j - 1 >= 0) dp[d][i][j] += dp[d][i][j - 1];
						if(i - 1 >= 0 && j - 1 >= 0) dp[d][i][j] -= dp[d][i - 1][j - 1];
					}
			for(int i = 0; i < q; i++){
				Integer[] txy = readInts(in.readLine());
				int t = txy[0], x1 = txy[1], y1 = txy[2], x2 = txy[3], y2 = txy[4];
				out.append(sum(x1, y1, x2, y2, dp[t % c])).append('\n');
			}
		}
		System.out.print(out);
	}

	public static Integer[] readInts(String line) {
		StringTokenizer st = new StringTokenizer(line.trim());
		Integer a[] = new Integer[st.countTokens()], index = 0;
		while (st.hasMoreTokens())
			a[index++] = Integer.parseInt(st.nextToken());
		return a;
	}

	public static Long[] readLongs(String line) {
		StringTokenizer st = new StringTokenizer(line.trim());
		Long a[] = new Long[st.countTokens()];
		int index = 0;
		while (st.hasMoreTokens())
			a[index++] = Long.parseLong(st.nextToken());
		return a;
	}

	public static Double[] readDoubles(String line) {
		StringTokenizer st = new StringTokenizer(line.trim());
		Double a[] = new Double[st.countTokens()];
		int index = 0;
		while (st.hasMoreTokens())
			a[index++] = Double.parseDouble(st.nextToken());
		return a;
	}
}
