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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in Actual solution is at the top
 *
 * @author MaxHeap
 */
public class Main {

  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    InputReader in = new InputReader(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    CStarSky solver = new CStarSky();
    solver.solve(1, in, out);
    out.close();
  }

  static class CStarSky {

    long[][][] sum;
    final static int MAX = 102;

    public void solve(int testNumber, InputReader in, PrintWriter out) {
      int n = in.nextInt(), q = in.nextInt(), c = in.nextInt();
      sum = new long[c + 1][MAX][MAX];
      for (int i = 0; i < n; ++i) {
        int x = in.nextInt(), y = in.nextInt(), s = in.nextInt();
        sum[s][x][y]++;
      }

      for (int i = 0; i < c + 1; ++i) {
        for (int x = 1; x < MAX; ++x) {
          for (int y = 1; y < MAX; ++y) {
            sum[i][x][y] += sum[i][x - 1][y] + sum[i][x][y - 1] - sum[i][x - 1][y - 1];
          }
        }
      }

      while (q-- > 0) {
        int curTime = in.nextInt();
        int x0 = in.nextInt(), y0 = in.nextInt(), x1 = in.nextInt(), y1 = in.nextInt();
        long ans = 0;
        for (int t = 0; t < c + 1; ++t) {
          int b = (curTime + t) % (c + 1);
          ans +=
              b * (sum[t][x1][y1] - sum[t][x0 - 1][y1] - sum[t][x1][y0 - 1] + sum[t][x0 - 1][y0
                  - 1]);
        }
        out.println(ans);
      }
    }

  }

  static class InputReader {

    private InputStream stream;
    private static final int DEFAULT_BUFFER_SIZE = 1 << 16;
    private static final int EOF = -1;
    private byte[] buf = new byte[DEFAULT_BUFFER_SIZE];
    private int curChar;
    private int numChars;

    public InputReader(InputStream stream) {
      this.stream = stream;
    }

    public int read() {
      if (this.numChars == EOF) {
        throw new UnknownError();
      } else {
        if (this.curChar >= this.numChars) {
          this.curChar = 0;

          try {
            this.numChars = this.stream.read(this.buf);
          } catch (IOException ex) {
            throw new InputMismatchException();
          }

          if (this.numChars <= 0) {
            return EOF;
          }
        }

        return this.buf[this.curChar++];
      }
    }

    public int nextInt() {
      int c;
      for (c = this.read(); isSpaceChar(c); c = this.read()) {
      }

      byte sgn = 1;
      if (c == 45) {
        sgn = -1;
        c = this.read();
      }

      int res = 0;

      while (c >= 48 && c <= 57) {
        res *= 10;
        res += c - 48;
        c = this.read();
        if (isSpaceChar(c)) {
          return res * sgn;
        }
      }

      throw new InputMismatchException();
    }

    public static boolean isSpaceChar(int c) {
      return c == 32 || c == 10 || c == 13 || c == 9 || c == EOF;
    }

  }
}

