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.

/**
 * @author: Mehul Raheja
 */
import java.util.*;
import java.io.*;

public class StarSky {

    /*
        Runtime = O()
     */
    static int N, M, K;
    static String s;
    static StringTokenizer st;
    static int[][] d;
    static int[][] d2;

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        d2 = new int[101][101];
        st = new StringTokenizer(br.readLine());
        int N = Integer.parseInt(st.nextToken());
        int Q = Integer.parseInt(st.nextToken());
        int C = Integer.parseInt(st.nextToken())+1;
        pair[] data = new pair[N];

        for (int i = 0; i < N; i++) {
            st = new StringTokenizer(br.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            int c = Integer.parseInt(st.nextToken());
            data[i] = new pair(a, b, c);
//            d[a][b] += 1;
//            d2[a][b] += c;
        }

        pair2[] data2 = new pair2[Q];
        for (int i = 0; i < Q; i++) {
            st = new StringTokenizer(br.readLine());
            int t = Integer.parseInt(st.nextToken());
            int x1 = Integer.parseInt(st.nextToken());
            int x2 = Integer.parseInt(st.nextToken());
            int y1 = Integer.parseInt(st.nextToken());
            int y2 = Integer.parseInt(st.nextToken());
            data2[i] = new pair2(t % C, x1, x2, y1, y2, i);
//            
//            int ans = query(d,x1,x2,y1,y2);
//            int ans2 = query(d,x1,x2,y1,y2);
//            System.out.println(ans + " " + ans2);
        }

        Arrays.sort(data2);

//        for (int i = 0; i < data2.length; i++) {
//            System.out.println(data2[i].a + " " + data2[i].b + " " + data2[i].c);
//        }
        
        int[] ans = new int[Q];

        int ind = 0;
        for (int T = 0; T < C; T++) {
            d = new int[101][101];
            for (pair e : data) {
                int a = e.a;
                int b = e.b;
                int c = e.c;
                d[a][b] += (c + T)%C;
            }

            
            for (int i = 1; i < d.length; i++) {
                for (int j = 1; j < d.length; j++) {
                    d[i][j] += d[i - 1][j] + d[i][j - 1] - d[i - 1][j - 1];
                }
            }
            //System.out.println("LOLQUERY " + T + " " + query(d,1,1,100,100));
            
            while(ind < data2.length && data2[ind].a == T){
                pair2 cur = data2[ind];
                int temp = query(d,cur.b,cur.c,cur.d,cur.e);
                ans[cur.f] = temp;
                ind++;
            }
        }
        
        //System.out.println(Arrays.toString(ans));
        for (int i = 0; i < ans.length; i++) {
            System.out.println(ans[i]);
        }
        //System.out.println(ans[ans.length-1]);

//        for (int i = 0; i < 10; i++) {
//            for (int j = 0; j < 10; j++) {
//                System.out.print(d[i][j]);
//            }
//            System.out.println();
//        }
//        
//        System.out.println(query(d,1,1,100,100));
    }

    public static int query(int[][] d, int x1, int y1, int x2, int y2) {
        x1--;
        y1--;
        int full = d[x2][y2];
        full -= d[x1][y2];
        full -= d[x2][y1];
        full += d[x1][y1];
        return full;
    }
}

class pair {

    int a, b, c;

    public pair(int _a, int _b, int _c) {
        this.a = _a;
        this.b = _b;
        this.c = _c;
    }

}

class pair2 implements Comparable<pair2> {

    int a, b, c, d, e, f;

    public pair2(int _a, int _b, int _c, int _d, int _e, int _f) {
        this.a = _a;
        this.b = _b;
        this.c = _c;
        this.d = _d;
        this.e = _e;
        this.f = _f;
    }

    @Override
    public int compareTo(pair2 t) {
        return (a == t.a) ? (b - t.b) : a - t.a;
    }

}
