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
#!/usr/bin/env python3
# created : 2020. 8. 19. 00:08

import os
from sys import stdin, stdout


def solve(tc):
    n, m, k = map(int, stdin.readline().split())
    seq = list(map(int, stdin.readline().split()))

    ops = []
    for i in range(m):
        ops.append(list(map(int, stdin.readline().split())))

    opsAcc = [0 for i in range(m+1)]
    for i in range(k):
        l, r = map(int, stdin.readline().split())
        opsAcc[l-1] += 1
        opsAcc[r] -= 1

    seqAcc = [0 for i in range(n+1)]
    seqAcc[ops[0][0]-1] += ops[0][2] * opsAcc[0]
    seqAcc[ops[0][1]] -= ops[0][2] * opsAcc[0]

    for i in range(1, m+1):
        opsAcc[i] += opsAcc[i-1]
        if i < m:
            seqAcc[ops[i][0]-1] += ops[i][2] * opsAcc[i]
            seqAcc[ops[i][1]] -= ops[i][2] * opsAcc[i]

    seq[0] += seqAcc[0]
    for i in range(1, n+1):
        seqAcc[i] += seqAcc[i-1]
        if i < n:
            seq[i] += seqAcc[i]

    print(*seq)


tcs = 1
tc = 1
while tc <= tcs:
    solve(tc)
    tc += 1
