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.io.*;
import java.util.*;

public class Assignment11 {

	public static void main(String[] args) throws IOException {
		final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		
		final int n = Integer.parseInt(st.nextToken());
		final int m = Integer.parseInt(st.nextToken());
		final int k = Integer.parseInt(st.nextToken());
		
		long[] arr = new long[n];
		st = new StringTokenizer(br.readLine());
		for(int i = 0; i < n; i++) {
			arr[i] = Integer.parseInt(st.nextToken());
		}
		
		int[][] ops = new int[m][5];
		
		for(int i = 0; i < m; i++) {
			st = new StringTokenizer(br.readLine());
			ops[i][0] = Integer.parseInt(st.nextToken())-1;
			ops[i][1] = Integer.parseInt(st.nextToken())-1;
			ops[i][2] = Integer.parseInt(st.nextToken());
		}
		
		for(int i = 0; i < k; i++) {
			st = new StringTokenizer(br.readLine());
			int x = Integer.parseInt(st.nextToken())-1;
			int y = Integer.parseInt(st.nextToken())-1;
			ops[x][3]++;
			ops[y][4]++;
		}
		
		long num = 0;
		long[] add = new long[n+1];
		
		for(int[] i: ops) {
			num += i[3];
			int l = i[0];
			int r = i[1];
			long d = i[2];
			add[l] += d*num;
			add[r + 1] += -d*num;
			num -= i[4];
		}
		
		long sum = 0;
		PrintWriter p = new PrintWriter(new BufferedOutputStream(System.out));
		for(int i = 0; i < n; i++) {
			sum += add[i];
			p.print((arr[i] + sum) + " ");
		}
		
		br.close();
		p.close();
	}

}
