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 scn = new Scanner(System.in);
	int n = scn.nextInt();
	int m = scn.nextInt();
	int k = scn.nextInt();
	long[] arr = new long[n+1];
	for(int i = 0; i < n; i++){
		arr[i] = scn.nextInt();
	}
	int[][] ops = new int[m+1][3];
	for(int i = 0; i < m; i++){
		ops[i][0] = scn.nextInt();
		ops[i][1] = scn.nextInt();
		ops[i][2] = scn.nextInt();
	}
	int[] count = new int[m+2];
	for(int i = 0; i < k; i++){
		int a = scn.nextInt();
		int b = scn.nextInt();
		count[a-1] += 1;
		count[b] -=1;
	}
	long[] adder = new long[n+2];
	long curr = 0;
	for (int i = 0; i < m;i++){
		curr += count[i];
		adder[ops[i][0]] +=  curr * ops[i][2];
		adder[ops[i][1] + 1] -=  curr * ops[i][2];
	}
	long alladd = 0;
	for (int i = 0; i < n; i++){
		alladd += adder[i+1];
		System.out.print(arr[i] + alladd + " ");
	}

	}
}