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 GregAndArray {

	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[] array = new long[n];
		for(int i = 0; i < n; ++i)
			array[i] = scan.nextInt();
		
		int[] ls = new int[m];
		int[] rs = new int[m];
		int[] ds = new int[m];
		for(int i = 0; i < m; ++i) {
			int l = scan.nextInt() - 1;
			int r = scan.nextInt() - 1;
			int d = scan.nextInt();
			ls[i] = l;
			rs[i] = r;
			ds[i] = d;
		}
		
		long[] ops = new long[m];
		
		//int size = 1;
		//while(size < m)
		//	size = size * 2;
		
		//int[] segTree = new int[size * 2 - 1];
		
		for(int i = 0; i < k; ++i) {
			int x = scan.nextInt() - 1;
			int y = scan.nextInt() - 1;
			ops[x] ++;
			if(y+1 < m)
				ops[y+1] --;
			//for(int j = x; j <= y; ++j)
			//	ops[j]+=1;
			//	updateValue(segTree, 0, n, j, 0);
			
		}
		
		long total = 0;
		for(int i = 0; i < m; ++i) {
			total += ops[i];
			ops[i] = total;
		}
			
		long[] change = new long[n];
		for(int i = 0; i < m; ++i) {
			change[ls[i]] += ds[i] * ops[i];
			if(rs[i]+1<n)
				change[rs[i]+1] -= ds[i] * ops[i];
			//for(int j = ls[i]; j <= rs[i]; ++j)
			//	array[j] += ds[i] * ops[i];
		}
		
		total = 0;
		for(int i = 0; i < n; ++i) {
			total += change[i];
			change[i] = total;
		}
		
		StringBuilder answer = new StringBuilder();
		
		for(int i = 0; i < n; ++i)
			answer.append((array[i] + change[i]) + " ");
		System.out.println(answer);
			//System.out.print(array[i] + " ");

	}
	
	public static void updateValue(int[] segTree, int ss, int se, int index, int si) {
		if(index < ss || index > se)
			return;
		
		segTree[si] += 1;
		if(ss != se) {
			int mid = (ss + se)/2;
			updateValue(segTree, ss, mid, index, 2 * si + 1);
			updateValue(segTree, mid + 1, se, index, 2 * si + 2);
		}
	}

}
