Greg has an array *a* = *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> and *m* operations. Each operation looks as: *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub>, *d*<sub class="lower-index">*i*</sub>, (1 ≤ *l*<sub class="lower-index">*i*</sub> ≤ *r*<sub class="lower-index">*i*</sub> ≤ *n*). To apply operation *i* to the array means to increase all array elements with numbers *l*<sub class="lower-index">*i*</sub>, *l*<sub class="lower-index">*i*</sub> + 1, ..., *r*<sub class="lower-index">*i*</sub> by value *d*<sub class="lower-index">*i*</sub>.

Greg wrote down *k* queries on a piece of paper. Each query has the following form: *x*<sub class="lower-index">*i*</sub>, *y*<sub class="lower-index">*i*</sub>, (1 ≤ *x*<sub class="lower-index">*i*</sub> ≤ *y*<sub class="lower-index">*i*</sub> ≤ *m*). That means that one should apply operations with numbers *x*<sub class="lower-index">*i*</sub>, *x*<sub class="lower-index">*i*</sub> + 1, ..., *y*<sub class="lower-index">*i*</sub> to the array.

Now Greg is wondering, what the array *a* will be after all the queries are executed. Help Greg.
The first line contains integers *n*, *m*, *k* (1 ≤ *n*, *m*, *k* ≤ 10<sup class="upper-index">5</sup>). The second line contains *n* integers: *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (0 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>) — the initial array.

Next *m* lines contain operations, the operation number *i* is written as three integers: *l*<sub class="lower-index">*i*</sub>, *r*<sub class="lower-index">*i*</sub>, *d*<sub class="lower-index">*i*</sub>, (1 ≤ *l*<sub class="lower-index">*i*</sub> ≤ *r*<sub class="lower-index">*i*</sub> ≤ *n*), (0 ≤ *d*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>).

Next *k* lines contain the queries, the query number *i* is written as two integers: *x*<sub class="lower-index">*i*</sub>, *y*<sub class="lower-index">*i*</sub>, (1 ≤ *x*<sub class="lower-index">*i*</sub> ≤ *y*<sub class="lower-index">*i*</sub> ≤ *m*).

The numbers in the lines are separated by single spaces.
from heapq import *
from queue import *
import sys, math, bisect
from collections import Counter, defaultdict
from math import *

input = lambda: sys.stdin.buffer.readline().decode().strip()
# print = sys.stdout.write
inl = lambda: list(map(int, input().split()))
intp = lambda: tuple(map(int, input().split()))


def solve():
    N, M, K = inl()
    A = inl()
    Operations = [inl() for _ in range(M)]

    # calculate how many times each operation is invoked
    psq = [0] * (K + 1)
    for _ in range(K):
        l, r = inl()
        psq[l - 1] += 1
        psq[r] -= 1
    for i in range(1, len(psq)):
        psq[i] += psq[i - 1]

    pso = [0] * (M + 1)
    for i, oper in enumerate(Operations):
        v = oper[2] * psq[i]
        pso[oper[0] - 1] += v
        pso[oper[1]] -= v

    for i in range(1, len(pso)):
        pso[i] += pso[i - 1]

    for i in range(N):
        A[i] += pso[i]
    print(*A)


for _ in range(int(1)):
    solve()
