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.*;
import java.io.*;


public class CodeForcesW10P1 {

    public static void main(String [] args) throws IOException  {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        String [] input1 = br.readLine().split(" ");
        int N = Integer.valueOf(input1[0]);
        int M = Integer.valueOf(input1[1]);
        int K = Integer.valueOf(input1[2]);

        int [] arr = new int[N+1];

        String [] input2 = br.readLine().split(" ");
        for (int i = 0; i < N; i ++){
            arr[i] = Integer.valueOf(input2[i]);
        }

        long [][] update_req = new long[M][3];
        for (int i = 0; i < M; i ++){
            String [] input3 = br.readLine().split(" ");
            update_req[i][0] = Long.valueOf(input3[0]);
            update_req[i][1] = Long.valueOf(input3[1]);
            update_req[i][2] = Long.valueOf(input3[2]);
        }

        long [] queryCache = new long[M];
        for (int i = 0; i < K; i ++){
            String [] input4 = br.readLine().split(" ");
            int start = Integer.valueOf(input4[0]) - 1;
            int end = Integer.valueOf(input4[1]);

            queryCache[start] += 1;

            if (end < queryCache.length){
                queryCache[end] -= 1;
            }
        }

        long totalCount = 0;
        for (int i = 0; i < M; i ++){
            queryCache[i] += totalCount;
            totalCount = queryCache[i];
            update_req[i][2] *= queryCache[i];
        }

        long [] updateCache = new long[N];
        for (int i = 0; i < M; i ++){
            int start = (int)update_req[i][0] - 1;
            int end = (int)update_req[i][1];

            updateCache[start] += update_req[i][2];

            if (end < updateCache.length){
                updateCache[end] -= update_req[i][2];
            }
        }

        totalCount = 0;
        for (int i = 0; i < N; i ++){
            updateCache[i] += totalCount;
            totalCount = updateCache[i];
        }

        for (int i = 0; i < N; i ++){
            long val = arr[i] + updateCache[i];
            System.out.print(val + " ");
        }
        System.out.println();
    }
}
