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
#include <bits/stdc++.h>
using namespace std;
long long N, M, K, l[100010], r[100010], d[100010], cnt[100010], a[100010],
    v[100010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> M >> K;
  int i;
  for (i = 1; i <= N; i++) cin >> a[i];
  for (i = 1; i <= M; i++) cin >> l[i] >> r[i] >> d[i];
  while (K--) {
    int x, y;
    cin >> x >> y;
    cnt[x]++, cnt[y + 1]--;
  }
  for (i = 1; i <= M; i++) {
    cnt[i] += cnt[i - 1];
    v[l[i]] += cnt[i] * d[i];
    v[r[i] + 1] -= cnt[i] * d[i];
  }
  for (i = 1; i <= N; i++) {
    v[i] += v[i - 1];
    cout << a[i] + v[i] << " ";
  }
  return 0;
}
