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
""" A. Greg and Array
"""
import sys

n, m, k = map(int, sys.stdin.readline().split())
arr = [x for x in map(int, sys.stdin.readline().split())]
ops = []
for x in range(m):
    ops.append((map(int, sys.stdin.readline().split())))
c = [0] * m # hold number of ops starting at i
for x in range(k):
    l, r = map(int, sys.stdin.readline().split())
    c[l - 1] += 1
    if r < m:
        c[r] -= 1 # query affects i - 1, stop at i
times_used = [0] * m # times each op is used
times_used[0] = c[0]
for x in range(1, m):
    times_used[x] = times_used[x - 1] + c[x]
c1 = [0] * n # hold number of updates starting at i
for x in range(m):
    l, r, d = ops[x]
    c1[l - 1] += times_used[x] * d
    if r < n:
        c1[r] -= times_used[x] * d
amount_added = [0] * n # amount to add to arr[i]
amount_added[0] = c1[0]
arr[0] += amount_added[0]
for x in range(1, n):
    amount_added[x] = amount_added[x - 1] + c1[x]
    arr[x] += amount_added[x]
print(*arr, sep = " ", end = "\n")
