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) throws IOException {

            boolean local = false;
            if (args.length > 0 && args[0].equals("LOCAL"))
                local = true;

            new Main(local).go();
        }

        boolean LOCAL;
        PrintWriter out;
        Reader in;
        BufferedReader br;

        Main(boolean local) throws IOException {

            this.LOCAL = local;

            try {

                //br = new BufferedReader( new FileReader("input.txt") );
                in = new Reader("input.txt");
                out = new PrintWriter( new BufferedWriter(new FileWriter("output.txt")) );
            }
            catch (Exception e) {

                //br = new BufferedReader( new InputStreamReader( System.in ) );
                in = new Reader();
                out = new PrintWriter( new BufferedWriter(new OutputStreamWriter(System.out)) );
            }
        }

        void go() throws IOException {

            int t = 1;

            if (this.LOCAL)
                t = in.nextInt();

            while (t > 0) {

                solve();
                t--;

                if (LOCAL)
                    out.println();

            }

            out.flush();
            out.close();
        }


        int inf = 2000000000;
        int mod = 1000000007;
        double eps = 1e-9;

        int n;
        int m;

        //ArrayList<Integer>[] g = new ArrayList[200010];

        void solve() throws IOException {

            int n = in.nextInt();
            int q = in.nextInt();
            int c = in.nextInt();

            int[][][] cnt = new int[11][101][101];

            for (int i = 0; i < n; i++) {
                int x = in.nextInt()-1;
                int y = in.nextInt()-1;
                int s = in.nextInt();
                cnt[s][x][y]++;
            }

            int[][][] sum = new int[11][101][101];

            for (int k = 0; k < c+1; k++)
                for (int i = 0; i < 101; i++) {

                    sum[k][i][0] += cnt[k][i][0];

                    for (int j = 1; j < 101; j++) {
                        sum[k][i][j] += sum[k][i][j - 1] + cnt[k][i][j];
                    }

                    if (i > 0)
                        for (int j = 0; j < 101; j++) {
                            sum[k][i][j] += sum[k][i - 1][j];
                        }
                }

            for (int i = 0; i < q; i++) {

                int t = in.nextInt();
                int x1 = in.nextInt()-1;
                int y1 = in.nextInt()-1;
                int x2 = in.nextInt()-1;
                int y2 = in.nextInt()-1;

                int ans = 0;

                for (int k = 0; k < c+1; k++) {

                    int cn = 0;

                    cn += sum[k][x2][y2];

                    if (x1 > 0) {
                        cn -= sum[k][x1 - 1][y2];
                    }

                    if (y1 > 0) {
                        cn -= sum[k][x2][y1 - 1];
                    }

                    if (x1 > 0 && y1 > 0) {
                        cn += sum[k][x1 - 1][y1 - 1];
                    }

                    ans += ((k + t)%(c+1))*cn;
                }

                out.println(ans);
            }
        }


        class Edge implements Comparable<Edge> {

            int a;
            int b;
            int c;
            int i;

            Edge(int a, int b, int c, int i) {

                this.a = a;
                this.b = b;
                this.c = c;
                this.i = i;
            }

            public int compareTo(Edge p) {

                if (c > p.c) return 1;
                if (c < p.c) return -1;
                if (i > p.i) return 1;
                if (i < p.i) return -1;
                return 0;
            }
        }

        class Pair implements Comparable<Pair>{

            int a;
            int b;

            Pair(int a, int b) {

                this.a = a;
                this.b = b;
            }

            public int compareTo(Pair p) {

    //            if (a > p.a) return 1;
    //            if (a < p.a) return -1;
                if (b > p.b) return 1;
                if (b < p.b) return -1;
                return 0;
            }
        }

        class Item {

            int a;
            int b;
            int c;

            Item(int a, int b, int c) {
                this.a = a;
                this.b = b;
                this.c = c;
            }

        }

        class Reader {

            BufferedReader  br;
            StringTokenizer tok;

            Reader(String file) throws IOException {
                br = new BufferedReader( new FileReader(file) );
            }

            Reader() throws IOException {
                br = new BufferedReader( new InputStreamReader(System.in) );
            }

            String next() throws IOException {

                while (tok == null || !tok.hasMoreElements())
                    tok = new StringTokenizer(br.readLine());
                return tok.nextToken();
            }

            int nextInt() throws NumberFormatException, IOException {
                return Integer.valueOf(next());
            }

            long nextLong() throws NumberFormatException, IOException {
                return Long.valueOf(next());
            }

            double nextDouble() throws NumberFormatException, IOException {
                return Double.valueOf(next());
            }


            String nextLine() throws IOException {
                return br.readLine();
            }

        }

    }