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 GregArray11_3 {
    public static void main(String[] args) throws IOException {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int m = scan.nextInt();
        int k = scan.nextInt();
        long[] arr = new long[n];
        int[][] ops = new int[m][3];
        for (int i = 0; i < n; i++) {
            arr[i] = scan.nextLong();
        }
        for (int i = 0; i < m; i++) {
            ops[i][0] = scan.nextInt() - 1;
            ops[i][1] = scan.nextInt() - 1;
            ops[i][2] = scan.nextInt();
        }
        int[] nTimesEachOp = new int[m + 1];    //+ & -
        for (int i = 0; i < k; i++) {
            int x = scan.nextInt() - 1;
            int y = scan.nextInt() - 1;
            nTimesEachOp[x]++;
            nTimesEachOp[y + 1]--;
//            for (int j = x; j <= y; j++) {
//                nTimesEachOp[j]++;
//            }
        }

        long[] update = new long[n + 1];
        long nTimesOp = 0;
        for (int op = 0; op < m; op++) {
            nTimesOp += nTimesEachOp[op];
//            for (int i = ops[op][0]; i <= ops[op][1]; i++) {
//                update[i] += (nTimesOp * ops[op][2]);
//            }
            update[ops[op][0]] += nTimesOp * ops[op][2];
            update[ops[op][1] + 1] -= nTimesOp * ops[op][2];
        }
        long updAmt = 0;
        for (int i = 0; i < n; i++) {
            updAmt += update[i];
            arr[i] += updAmt; //read the + & -
            System.out.print(arr[i] + " ");
        }
    }
}
