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;
const int maxn = 100000 + 100;
long long l[maxn], r[maxn], ezafe[maxn], chand[maxn], d[maxn], a[maxn];
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < m; i++) {
    cin >> l[i] >> r[i] >> d[i];
    l[i]--;
    r[i]--;
  }
  for (long long i = 0; i < k; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    chand[x]++;
    chand[y + 1]--;
  }
  for (long long i = 1; i < maxn; i++) chand[i] += chand[i - 1];
  for (long long i = 0; i < m; i++) {
    ezafe[l[i]] += ((long long)chand[i] * (long long)d[i]);
    ezafe[r[i] + 1] -= ((long long)chand[i] * (long long)d[i]);
  }
  for (long long i = 1; i < n; i++) ezafe[i] += ezafe[i - 1];
  for (long long i = 0; i < n; i++) cout << (a[i] + ezafe[i]) << ' ';
}
