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 Program {

    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;

    static final int MOD = 1000000007;
    static final double INF = 10000000000.0;
    
    
	void solve() throws IOException {
		int n, m, k;
		n = nextInt();
		m = nextInt();
		k = nextInt();
		int[] a, l, r, d;
		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] = nextInt();
		for(int i = 0; i < m; i++)
		{
			l[i] = nextInt();
			r[i] = nextInt();
			d[i] = nextInt();
		}
		long[] dp1, dp2;
		dp1 = new long[m+2];
		dp2 = new long[n+2];
		for(int i = 0; i < k; i++)
		{
			dp1[nextInt()-1]++;
			dp1[nextInt()]--;
		}
		long cnt = 0;
		for(int i = 0; i < m; i++)
		{
			cnt += dp1[i];
			dp2[l[i]-1] += d[i] * cnt;
			dp2[r[i]] -= d[i]*cnt;
		}
		cnt = 0;
		for(int i = 0; i < n; i++)
		{
			cnt+=dp2[i];
			out.print((a[i]+cnt) + " ");
		}
	}

    Program() throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }

    public static void main(String[] args) throws IOException {
        new Program();
    }

    String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }

    String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}