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.util.Scanner;

/**
 * Created by Alex on 4/17/17.
 * http://codeforces.com/problemset/problem/295/A
 */
public class WB_295A {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        int k = in.nextInt();
        int[] arr = new int[n + 1];
        for (int i = 1; i <= n; i++) {
            arr[i] = in.nextInt();
        }

        // All operations
        Operation[] ops = new Operation[m + 1];
        for (int i = 1; i <= m; i++) {
            ops[i] = new Operation(in.nextInt(), in.nextInt(), in.nextInt());
        }

        // Changes of number of times each operation is performed
        int[] numOps = new int[m + 2];
        for (int i = 0; i < k; i++) {
            int start = in.nextInt();
            int end = in.nextInt();
            numOps[start]++;
            numOps[end + 1]--;
        }

        // Changes of the value to be added to each element of original array
        long[] values = new long[n + 2];
        long numOp = 0; // running total for num times each op is performed
        for (int i = 1; i <= m; i++) {
            numOp += numOps[i];
            values[ops[i].left] += numOp * ops[i].value;
            values[ops[i].right + 1] -= numOp * ops[i].value;
        }

        StringBuilder ans = new StringBuilder();
        long value = 0; // running total for value to be added to each element
        for (int i = 1; i <= n; i++) {
            value += values[i];
            ans.append(arr[i] + value).append(" ");
        }

        System.out.println(ans.toString());
    }

    private static class Operation {
        int left, right, value;

        Operation(int left, int right, int value) {
            this.left = left;
            this.right = right;
            this.value = value;
        }
    }
}
