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 Solution {
    
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int length = in.nextInt();
        int ops = in.nextInt();
        int queries = in.nextInt();
        long[] arr = new long[length+1];
        in.nextLine();
        for (int i = 0; i < length; i++){
            arr[i] = in.nextLong();
        }
        int[][] opers = new int[ops + 1][3];//0:L, 1:R, 2:D
        for (int i = 0; i < ops; i++){
            in.nextLine();
            opers[i][0] = in.nextInt();
            opers[i][1] = in.nextInt();
            opers[i][2] = in.nextInt();
        }
        int[][] q = new int[queries + 1][2];//0:start, 1:end
        for (int i = 0; i < queries; i++){
            in.nextLine();
            q[i][0] = in.nextInt();
            q[i][1] = in.nextInt();
        }
        int[] multi = new int[ops + 1];
        for (int i = 0; i < queries; i++){//multiply relative to previous
            multi[q[i][0]-1]++;
            multi[q[i][1]]--;
        }
        
        long[] add = new long[length+1];
        long curr_mult = 0;
        for (int i = 0; i < ops; i++){
            curr_mult += multi[i];
            long newD = curr_mult * opers[i][2];
            add[opers[i][0]-1] += newD; //num to add relative to previous
            add[opers[i][1]] -= newD;
        }
        long curr_add = 0;
        for (int i = 0; i < length; i++){
            curr_add += add[i];
            arr[i] += curr_add;
            System.out.print(arr[i] + " ");
        }
    }
    
}