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.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class _A9_Star_Sky
{
	public static void main(String[] args)
	{
		int T = 1;
		// int T = s.nextInt();
		while (T > 0)
		{
			// System.out.println("TEST CASE: "+T);
			solve();
			T--;
		}
	}

	static class FastReader
	{
		BufferedReader br;
		StringTokenizer st;

		public FastReader()
		{
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String next()
		{
			while (st == null || !st.hasMoreElements())
			{
				try
				{
					st = new StringTokenizer(br.readLine());
				} catch (IOException e)
				{
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt()
		{
			return Integer.parseInt(next());
		}

		long nextLong()
		{
			return Long.parseLong(next());
		}

		double nextDouble()
		{
			return Double.parseDouble(next());
		}

		String nextLine()
		{
			String str = "";
			try
			{
				str = br.readLine();
			} catch (IOException e)
			{
				e.printStackTrace();
			}
			return str;
		}
	}

	static FastReader s = new FastReader();

	public static void solve()
	{
		int n = s.nextInt();
		int q = s.nextInt();
		int mx = s.nextInt() + 1;
		int[][][] arr = new int[mx][101][101];
		for (int i = 0; i < n; i++)
		{
			int x = s.nextInt();
			int y = s.nextInt();
			int b = s.nextInt();
			arr[b][x][y]++;
		}
		for (int i = 0; i < mx; i++)
		{
			for (int j = 1; j < 101; j++)
			{
				for (int k = 1; k < 101; k++)
				{
					arr[i][j][k] = arr[i][j][k] + arr[i][j - 1][k] + arr[i][j][k - 1] - arr[i][j - 1][k - 1];
				}
			}
		}
		for (int i = 0; i < q; i++)
		{
			long t = s.nextLong();
			int x1 = s.nextInt();
			int y1 = s.nextInt();
			int x2 = s.nextInt();
			int y2 = s.nextInt();
			long ans = 0l;
			for (int j = 0; j < mx; j++)
			{
				long bright = ((j + t) % (mx));
				long amt = arr[j][x2][y2] + arr[j][x1 - 1][y1 - 1] - arr[j][x1 - 1][y2] - arr[j][x2][y1 - 1];
				ans += (bright * amt);
			}
			System.out.println(Math.max(0, ans));
		}
	}
}