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;
template <typename T>
using vc = vector<T>;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> ar(n + 1);
  for (int i = 0; i < n; i++) cin >> ar[i];
  vc<vc<ll>> que(m, vc<ll>(3));
  for (int i = 0; i < m; i++) {
    cin >> que[i][0] >> que[i][1] >> que[i][2];
    que[i][0]--;
    que[i][1]--;
  }
  int x, y;
  vc<ll> op(m);
  while (k--) {
    cin >> x >> y;
    x--;
    op[x]++;
    if (y < m) op[y]--;
  }
  for (int i = 1; i < m; i++) op[i] += op[i - 1];
  vc<ll> res(n);
  for (int i = 0; i < m; i++) {
    int st = que[i][0];
    int en = que[i][1];
    res[st] += op[i] * que[i][2];
    if (en + 1 < n) res[en + 1] -= op[i] * que[i][2];
  }
  for (int i = 1; i < n; i++) res[i] += res[i - 1];
  for (int i = 0; i < n; i++) cout << res[i] + ar[i] << ' ';
}
