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

/**
 * @author Don Li
 */
public class StarSky {
    
    int N = 105;
    
    void solve() {
        int n = in.nextInt(), q = in.nextInt(), c = in.nextInt();
        int[] x = new int[n], y = new int[n], s = new int[n];
        for (int i = 0; i < n; i++) {
            x[i] = in.nextInt();
            y[i] = in.nextInt();
            s[i] = in.nextInt();
        }
        
        int[][][] b = new int[c + 1][N][N];
        for (int p = 0; p <= c; p++) {
            for (int i = 0; i < n; i++) {
                b[p][x[i]][y[i]] += (s[i] + p) % (c + 1);
            }
        }
        
        int[][][] sum = new int[c + 1][N + 1][N + 1];
        for (int p = 0; p <= c; p++) {
            for (int u = 1; u < N; u++) {
                for (int v = 1; v < N; v++) {
                    sum[p][u + 1][v + 1] = sum[p][u][v + 1] + sum[p][u + 1][v] - sum[p][u][v] + b[p][u][v];
                }
            }
        }
        
        for (int i = 0; i < q; i++) {
            int t = in.nextInt(), x1 = in.nextInt(), y1 = in.nextInt(), x2 = in.nextInt(), y2 = in.nextInt();
            t %= c + 1;
            int ans = sum[t][x2 + 1][y2 + 1] - sum[t][x1][y2 + 1] - sum[t][x2 + 1][y1] + sum[t][x1][y1];
            out.println(ans);
        }
    }
    
    public static void main(String[] args) {
        in = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(System.out);
        new StarSky().solve();
        out.close();
    }
    
    static FastScanner in;
    static PrintWriter out;
    
    static class FastScanner {
        BufferedReader in;
        StringTokenizer st;
        
        public FastScanner(BufferedReader in) {
            this.in = in;
        }
        
        public String nextToken() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(in.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        public int nextInt() {
            return Integer.parseInt(nextToken());
        }
        
        public long nextLong() {
            return Long.parseLong(nextToken());
        }
        
        public double nextDouble() {
            return Double.parseDouble(nextToken());
        }
    }
}
