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.*;

/**
 * Problem Cf835C
 */
public class Cf835C {

    public void run() {
        int N = in.nextInt();
        int Q = in.nextInt();
        int C = in.nextInt();
        Map<Integer, Map<Integer, List<Integer>>> XY = new HashMap<>();
        for(int i = 0; i < N; ++i){
            int x = in.nextInt();
            int y = in.nextInt();
            int s = in.nextInt();
            XY.computeIfAbsent(x, key -> new HashMap<>())
              .computeIfAbsent(y, key -> new ArrayList<>())
              .add(s);
        }
        int[][][] B = new int[101][101][11];
        for(int c = 0; c <= C; ++c){
            for(int x = 1; x <= 100; ++x){
                for(int y = 1; y <= 100; ++y){
                    int b = 0;
                    for(int s : get(XY,x,y)){
                        b+=(s+c)%(C+1);
                    }
                    B[x][y][c] = B[x-1][y][c] + B[x][y-1][c] - B[x-1][y-1][c] + b;
                }
            }
        }
        while(Q-->0){
            int t = in.nextInt()%(C+1);
            int x1 = in.nextInt();
            int y1 = in.nextInt();
            int x2 = in.nextInt();
            int y2 = in.nextInt();
            int b = B[x2][y2][t] - B[x2][y1-1][t] - B[x1-1][y2][t] + B[x1-1][y1-1][t];
            out.println(b);
        }

    }

    List<Integer> get(Map<Integer, Map<Integer, List<Integer>>> XY, int x, int y){
        return Optional.of(XY)
                .map(map -> map.get(x))
                .map(map -> map.get(y))
                .orElse(Collections.emptyList());
    }


    /***********************************************************
     *                      BOILERPLATE                        *
    /***********************************************************/
    public InputReader in = new InputReader(System.in);
    public PrintWriter out = new PrintWriter(System.out);
    public void close() {out.close();}
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokens = null;
        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);}
        private boolean prime() {
            while (tokens == null || !tokens.hasMoreTokens()) {
                try {
                    String line = reader.readLine();
                    if(line==null) return false; // EOF
                    tokens = new StringTokenizer(line);
                } catch (IOException e) {throw new RuntimeException(e);}
            } return true;}
        public boolean hasNext() {return prime();}
        public String next() {prime(); return tokens.nextToken();}
        public int nextInt() {return Integer.parseInt(next());}
        public long nextLong() {return Long.parseLong(next());}
        public double nextDouble() {return Double.parseDouble(next());}
    }
    public static void main(String[] args) {
        Cf835C task = new Cf835C(); task.run(); task.close();}
}
