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 ladder{

	
	public static void main(String[] args) throws IOException {
		Scanner sc=new Scanner();
		PrintWriter out=new PrintWriter(System.out);
		int n=sc.nextInt(),m=sc.nextInt(),k=sc.nextInt();
		int []a=new int [n],l=new int [m],r=new int [m],d=new int [m];
		for(int i=0;i<n;i++)
			a[i]=sc.nextInt();
		for(int i=0;i<m;i++)
		{
			l[i]=sc.nextInt()-1;
			r[i]=sc.nextInt()-1;
			d[i]=sc.nextInt();
		}
		int []times=new int [m+1];
		while(k-->0)
		{
			int x=sc.nextInt()-1,y=sc.nextInt();
			times[x]++;
			times[y]--;
		}
		long []prefix=new long [n+1];
		long t=0;
		for(int i=0;i<m;i++)
		{
			t+=times[i];
			long v=t*d[i];
			prefix[l[i]]+=v;
			prefix[r[i]+1]-=v;
		}
		long pre=0;
		for(int i=0;i<n;i++)
		{
			pre+=prefix[i];
			out.println(a[i]+pre);
		}
		out.close();

	}
	static class Scanner
	{
		BufferedReader br;
		StringTokenizer st;
		Scanner(){
			br=new BufferedReader(new InputStreamReader(System.in));
		}
		Scanner(String fileName) throws FileNotFoundException{
			br=new BufferedReader(new FileReader(fileName));
		}
		String next() throws IOException {
			while(st==null || !st.hasMoreTokens())
				st=new StringTokenizer(br.readLine());
			return st.nextToken();
		}
		String nextLine() throws IOException {
			return br.readLine();
		}
		int nextInt() throws IOException{
			return Integer.parseInt(next());
		}
		long nextLong()  throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}
		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}
	}
}
