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.io.PrintWriter;
import java.util.StringTokenizer;

public class Div2_427C {

	public static void main(String[] args) throws IOException {
		BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter printer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		StringTokenizer inputData = new StringTokenizer(reader.readLine());
		int numP = Integer.parseInt(inputData.nextToken());
		int numQ = Integer.parseInt(inputData.nextToken());
		int mod = Integer.parseInt(inputData.nextToken()) + 1;

		int[][][] grid = new int[mod][101][101];

		int[][][] preS = new int[mod][101][101];

		for (int i = 0; i < numP; i++) {
			inputData = new StringTokenizer(reader.readLine());
			int x = Integer.parseInt(inputData.nextToken());
			int y = Integer.parseInt(inputData.nextToken());
			int s = Integer.parseInt(inputData.nextToken());

			for (int t = 0; t < mod; t++) {
				grid[t][y][x] += (s + t) % mod;
			}
		}

		for (int res = 0; res < mod; res++) {
			for (int y = 1; y <= 100; y++) {
				for (int x = 1; x <= 100; x++) {
					preS[res][y][x] = preS[res][y][x - 1] + grid[res][y][x];
				}
			}
		}

		while (numQ-- > 0) {
			inputData = new StringTokenizer(reader.readLine());
			int t = Integer.parseInt(inputData.nextToken()) % mod;
			int left = Integer.parseInt(inputData.nextToken());
			int low = Integer.parseInt(inputData.nextToken());
			int right = Integer.parseInt(inputData.nextToken());
			int high = Integer.parseInt(inputData.nextToken());
			int sum = 0;

			for (int y = low; y <= high; y++) {
				sum += preS[t][y][right] - preS[t][y][left - 1];
			}
			printer.println(sum);
		}
		printer.close();
	}

}
