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.*;
import java.util.*;
 
 
public class Main{
   public static void main(String[] args) {
      MyScanner s = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
      
      // Start writing your solution here. -------------------------------------
   
      /*
      int n      = sc.nextInt();        // read input as integer
      long k     = sc.nextLong();       // read input as long
      double d   = sc.nextDouble();     // read input as double
      String str = sc.next();           // read input as String
      String s   = sc.nextLine();       // read whole line as String

      int result = 3*n;
      out.println(result);                    // print via PrintWriter
      */

      // Stop writing your solution here. -------------------------------------
      
            int n = s.nextInt();
      int q = s.nextInt();
      int c = s.nextInt();
      int count[][][] = new int[11][101][101];
      int dp[][][] = new int[11][101][101];
      for(int i = 0; i < n; i++){
        int x = s.nextInt();
        int y = s.nextInt();
        int b = s.nextInt();
        count[b][x][y]++;
      }
      for(int i = 1; i <= 100; i++){
        for(int j = 1; j <= 100; j++){
          for(int p = 0; p <= c; p++){
            dp[p][i][j] = dp[p][i-1][j] + dp[p][i][j-1] - dp[p][i-1][j-1] + count[p][i][j];
          }
        }
      }
      while(q-- > 0){
        int t = s.nextInt();
        int x1 = s.nextInt();
        int y1 = s.nextInt();
        int x2 = s.nextInt();
        int y2 = s.nextInt();
        int res = 0;
        for(int p = 0; p <= c; p++){
          	int bright = (p+t)%(c+1);
          int stars = dp[p][x2][y2] - dp[p][x1-1][y2] - dp[p][x2][y1-1] + dp[p][x1-1][y1-1];
          res += bright*stars;
        }
        System.out.println(res);
      }
      
      out.close();
   }

     

   //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;
      
   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         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;
      }

   }
   //--------------------------------------------------------
}