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 Codeforce {
	
	private static final int N = 100005;
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		int k = sc.nextInt();
		long[] a = new long[N];
		long[] lz = new long[N];
		long[] lzo = new long[N];
		long[] d = new long[N];
		
		int[] l = new int[N];
		int[] r = new int[N];
		for (int i = 1; i <= n; i++) {
			a[i] = sc.nextInt();
		}
		for (int i = 1; i <= m; i++) {
			l[i] = sc.nextInt();
			r[i] = sc.nextInt();
			d[i] = sc.nextInt();
		}
		for (int i = 0; i < k; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			lzo[x]++;
			lzo[y + 1]--;
		}
		
		long sum = 0;
		for (int i = 1; i <= m; i++) {
			sum += lzo[i];
			lz[(int) l[i]] += d[i] * sum;
			lz[(int) r[i] + 1] -= d[i] * sum;
		}
		sum = 0;
		for (int i = 1; i <= n; i++) {
			sum += lz[i]; 
			System.out.print(a[i] + sum + " ");
		}
	}
}