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;

public class p295A {

	public static void main(String[] args) {
		Scanner s = new Scanner (System.in);
		int size = s.nextInt();
		int numOp = s.nextInt();
		int numQue = s.nextInt();
		
		int arr[] = new int[100000];
		int l[] = new int[100000];
		int r[] = new int[100000];
		int d[] = new int[100000];
		int queue[] = new int[100001];
		long comp[] = new long[100001];
		
		for (int i = 0; i < size; ++i) {
			arr[i] = s.nextInt();
		}
		
		for (int i = 0; i < numOp; ++i) {
			l[i] = s.nextInt();
			r[i] = s.nextInt();
			d[i] = s.nextInt();
		}
		
		for (int i =0; i < numQue; ++i) {
			++queue[s.nextInt()-1];
			--queue[s.nextInt()];
		}
		
		long c = 0;
		for (int i = 0; i < numOp; ++i) {
			c += queue[i];
			long val = c * d[i];
			comp[l[i]-1] += val;
			comp[r[i]] -= val;
		}
		
		c = 0;
		for (int i = 0; i < size; ++i) {
			c += comp[i];
			System.out.print(c+arr[i]);
			if (i+1 < size) 
				System.out.print(" ");
		}
		System.out.println();
	}

}
