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.*;
import java.io.*;

public class CodeForces {

    public static void main(String[] args) throws IOException {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        int m = input.nextInt();
        int k = input.nextInt();
        int[] arr = new int[n];

        for(int i = 0; i < n; i++)
            arr[i] = input.nextInt();
        int[] l = new int[m];
        int[] r = new int[m];
        int[] d = new int[m];
        for(int i = 0; i < m; i++) {
            l[i] = input.nextInt();
            r[i] = input.nextInt();
            d[i] = input.nextInt();
        }
        long[] ops = new long[m + 1];
        for(int i = 0; i < k; i++) {
            int x = input.nextInt();
            int y = input.nextInt();
            ops[x - 1]++;
            ops[y]--;
        }
        long[] res = new long[n + 1];
        long count = 0;
        for(int i = 0; i < m; i++) {
            count += ops[i];
            long newVal = count * d[i];
            res[l[i] - 1] += newVal;
            res[r[i]] -= newVal;
        }
        count = 0;
        for(int i = 0; i < n; i++) {
            count += res[i];
            System.out.println(count + arr[i]);
        }
    }
}
