Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value di.

Greg wrote down k queries on a piece of paper. Each query has the following form: xi, yi, (1 ≤ xi ≤ yi ≤ m). That means that one should apply operations with numbers xi, xi + 1, ..., yi to the array.

Now Greg is wondering, what the array a will be after all the queries are executed. Help Greg.

Input

The first line contains integers n, m, k (1 ≤ n, m, k ≤ 105). The second line contains n integers: a1, a2, ..., an (0 ≤ ai ≤ 105) — the initial array.

Next m lines contain operations, the operation number i is written as three integers: li, ri, di, (1 ≤ li ≤ ri ≤ n), (0 ≤ di ≤ 105).

Next k lines contain the queries, the query number i is written as two integers: xi, yi, (1 ≤ xi ≤ yi ≤ m).

The numbers in the lines are separated by single spaces.

Output

On a single line print n integers a1, a2, ..., an — the array after executing all the queries. Separate the printed numbers by spaces.

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams of the %I64d specifier.

Examples

Input

3 3 3
1 2 3
1 2 1
1 3 2
2 3 4
1 2
1 3
2 3


Output

9 18 17


Input

1 1 1
1
1 1 1
1 1


Output

2


Input

4 3 6
1 2 3 4
1 2 1
2 3 2
3 4 4
1 2
1 3
2 3
1 2
1 3
2 3


Output

5 18 31 20
import java.io.*;
import java.util.StringTokenizer;

public class Main {
    final static boolean CONTEST = true;
    //region shit

    BufferedReader cin;
    PrintWriter cout;
    StringTokenizer stringTokenizer;

    public static void main(String[] args) throws Exception {
        Main sol = new Main();
        if (CONTEST) {
            sol.cin = new BufferedReader(new InputStreamReader(System.in));
        } else {
            sol.cin = new BufferedReader(new FileReader("input.txt"));
        }
        sol.cout = new PrintWriter(System.out);
        sol.solve();
        sol.cin.close();
        sol.cout.close();
    }

    StringTokenizer tokenizer;

    private void readLine() throws IOException {
        tokenizer = new StringTokenizer(cin.readLine());
    }

    private String nextToken() throws Exception {
        return tokenizer.nextToken();
    }

    private int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    long[] summation(int[] a) {
        int n = a.length - 1;
        long[] result = new long[n];
        result[0] = a[0];
        for (int i = 1; i < n; i++) {
            result[i] = result[i - 1] + a[i];
        }
        return result;
    }

    class Aggregator {

        private long[] increments;

        private int n;

        Aggregator(int capacity) {
            this.n = capacity;
            increments = new long[capacity + 1];
        }

        public void query(int l, int r) {
            query(l, r, 1);
        }

        public void query(int l, int r, long d) {
            increments[l] += d;
            increments[r + 1] -= d;
        }

        long[] release() {

            long[] result = new long[n];
            result[0] = increments[0];
            for (int i = 1; i < n; i++) {
                result[i] = result[i - 1] + increments[i];
            }
            return result;
        }
    }

    private void solve() throws Exception {
        readLine();
        int n = nextInt();
        int m = nextInt();
        int k = nextInt();
        readLine();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
        }
        int[] l = new int[m];
        int[] r = new int[m];
        int[] d = new int[m];
        for (int i = 0; i < m; i++) {
            readLine();
            l[i] = nextInt() - 1;
            r[i] = nextInt() - 1;
            d[i] = nextInt();
        }
        Aggregator repetitionsAgr = new Aggregator(m);
        for (int i = 0; i < k; i++) {
            readLine();
            repetitionsAgr.query(nextInt() - 1, nextInt() - 1);
        }
        long[] repetitions = repetitionsAgr.release();
        Aggregator additionsAgr = new Aggregator(n);
        for (int i = 0; i < m; i++) {
            additionsAgr.query(l[i], r[i], d[i] * repetitions[i]);
        }
        long[] additions = additionsAgr.release();
        for (int i = 0; i < n; i++) {
            cout.print((a[i] + additions[i]) + " ");
        }

    }
}