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.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class C427 {
    public static void main(String[] args) {
        FS scan = new FS(System.in);
        int N = scan.nextInt(), Q = scan.nextInt(), C = scan.nextInt();
        int[][][] mat = new int[C+1][100][100];
        
        for(int i=0;i<N;i++){
            int x = scan.nextInt()-1, y = scan.nextInt()-1;
            int c = scan.nextInt();
            
            for(int j=0;j<=C;j++){
                mat[j][x][y]+=c;
                c = (c+1)%(C+1);
            }
        }
        
       
        
        int[][][] sums = new int[C+1][][];
        for(int i=0;i<=C;i++){
            sums[i] = sum(mat[i]);
        }
       
        PrintWriter out = new PrintWriter(System.out);
        for(int i=0;i<Q;i++){
            int t = scan.nextInt(), r1 = scan.nextInt(), c1 = scan.nextInt(), r2 = scan.nextInt(), c2 = scan.nextInt();
            
            t%=(C+1);
            out.println(query(sums[t], r1,c1,r2,c2));
        }
        out.close();
        
    }
    
    private static int[][] sum(int[][] mat) {
        int[][] sum = new int[mat.length+1][mat[0].length+1];
        for(int i=1;i<sum.length;i++){
            for(int j=1;j<sum[0].length;j++){
                sum[i][j] = sum[i][j-1]+mat[i-1][j-1];
            }
        }
        for(int j=1;j<sum[0].length;j++){
            for(int i=1;i<sum.length;i++){
                sum[i][j] += sum[i-1][j];
            }
        }
        return sum;
    }
    
    private static int query(int[][] mat, int r1, int c1, int r2, int c2) {
        return mat[r2][c2]-mat[r2][c1-1]-mat[r1-1][c2]+mat[r1-1][c1-1];
    }
    
    
    private static class FS {
        BufferedReader br;
        StringTokenizer st;
        public FS(InputStream in) {
            br = new BufferedReader(new InputStreamReader(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());}
    }
}
