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;
using ll = long long;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  ll a, b, val, n, m, k, crt1 = 0, crt = 0;
  cin >> n >> m >> k;
  vector<ll> v(n + 2), times(m + 2), init(n + 2);
  for (int i = 1; i <= n; ++i) cin >> init[i];
  vector<tuple<int, int, int>> q;
  q.emplace_back(0, 0, 0);
  for (int i = 0; i < m; ++i) {
    cin >> a >> b >> val;
    q.emplace_back(a, b, val);
  }
  for (int i = 0; i < k; ++i) {
    cin >> a >> b;
    times[a]++;
    times[b + 1]--;
  }
  for (int i = 1; i <= m; ++i) {
    crt1 += times[i];
    tie(a, b, val) = q[i];
    v[a] += crt1 * val;
    v[b + 1] -= crt1 * val;
  }
  for (int i = 1; i <= n; ++i) {
    crt += v[i];
    cout << init[i] + crt << ' ';
  }
  return 0;
}
