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.*;

public class GregAndArray {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        int m = cin.nextInt();
        int k = cin.nextInt();
        point1 array[] = new point1[n];
        point operations[] = new point[m];
        int array1[] = new int[m];
        for (int i = 0; i < n; i++) {
            array[i] = new point1();
            array[i].x = cin.nextLong();
            array[i].y = 0;
        }
        for (int i = 0; i < m; i++) {
            operations[i] = new point();
            operations[i].x = cin.nextInt();
            operations[i].y = cin.nextInt();
            operations[i].z = cin.nextInt();
        }
        for (int i = 0; i < k; i++) {
            int a = cin.nextInt();
            int b = cin.nextInt();
            array1[a - 1] += 1;
            if (b != m)
                array1[b] -= 1;
        }
        long cnt = 0;
        for (int i = 0; i < m; i++) {
            array1[i] += cnt;
            cnt = array1[i];
        }
        cnt = 0;
        for (int i = 0; i < m; i++) {
            cnt = (long) array1[i] * operations[i].z;
            array[operations[i].x - 1].y += cnt;
            if (operations[i].y != n)
                array[operations[i].y].y -= cnt;
        }
        cnt = 0;
        for (int i = 0; i < n; i++) {
            cnt += array[i].y;
            array[i].x += cnt;
        }
        StringBuilder str = new StringBuilder();
        for (int i = 0; i < n; i++) {
            str.append(array[i].x + " ");
        }
        System.out.print(str);
        cin.close();
    }
    static class point {
        int x;
        int y;
        int z;
    }
    static class point1 {
        long x;
        long y;
    }
}