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.*;
public class Greg {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		int m = scan.nextInt();
		int k = scan.nextInt();
		long arr[] = new long[n];
		for(int i = 0; i <n; i++) {
			arr[i] = scan.nextInt();
		}
		int ops[][] = new int[m][3];
		for(int j = 0; j < m; j++) {
			ops[j][0] = scan.nextInt();
			ops[j][1] = scan.nextInt();
			ops[j][2] = scan.nextInt();
		}
		long up[] = new long[m+1];
		long up2[] = new long[n+1];
		for(int i=0; i<k;i++) {
			int x = scan.nextInt();
			int y = scan.nextInt();
			up[x-1]++;
			up[y]--;
		}
		long sum = 0;
		for(int i = 0; i < m; i++) {
			sum+=up[i];
			up2[ops[i][0]-1]+=ops[i][2]*sum;
			up2[ops[i][1]]-=ops[i][2]*sum;
		}
		sum = 0;
		for(int i=0;i<n;i++) {
			sum+=up2[i];
			arr[i]+=sum;
		}
		for(int i=0; i<n;i++) {
			System.out.print(arr[i]+ " ");
		}
		System.out.println();
	}

}
