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 = 1e5 + 7;
long long mas[maxn] = {0};
long long a[maxn];
long long mean[maxn] = {0};
long long ask[maxn][3];
long long ans[maxn] = {0};
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> ask[i][0] >> ask[i][1] >> ask[i][2];
  for (int i = 1; i <= k; i++) {
    int l, r;
    cin >> l >> r;
    mas[l]++;
    mas[r + 1]--;
    continue;
  }
  for (int i = 1; i <= m; i++) mas[i] += mas[i - 1];
  for (int i = 1; i <= m; i++) {
    ans[ask[i][0]] += ask[i][2] * mas[i];
    ans[ask[i][1] + 1] -= ask[i][2] * mas[i];
    continue;
  }
  for (int i = 1; i <= n; i++) {
    ans[i] += ans[i - 1];
    cout << a[i] + ans[i] << " ";
    continue;
  }
  cout << endl;
  return 0;
}
