#include <bits/stdc++.h>
using namespace std;
long long int a[100100];
long long int l[100100], r[100100], d[100100], oper[100100], b[100100];
int main() {
  pair<long long int, long long int> p[100100];
  pair<long long int, long long int> op[100100];
  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 >> l[i] >> r[i] >> d[i];
  }
  int x, y;
  for (int i = 1; i <= k; i++) {
    cin >> x >> y;
    op[x].first++;
    op[y].second++;
  }
  long long int xx = 0;
  for (int i = 1; i <= m; i++) {
    xx += op[i].first;
    oper[i] = xx;
    xx -= op[i].second;
  }
  xx = 0;
  for (int i = 1; i <= m; i++) {
    p[l[i]].first += (d[i] * oper[i]);
    p[r[i]].second += (d[i] * oper[i]);
  }
  for (int i = 1; i <= n; i++) {
    xx += p[i].first;
    b[i] = xx;
    b[i] += a[i];
    cout << b[i] << " ";
    xx -= p[i].second;
  }
}

### Prompt

Your challenge is to write a cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int a[100100];
long long int l[100100], r[100100], d[100100], oper[100100], b[100100];
int main() {
  pair<long long int, long long int> p[100100];
  pair<long long int, long long int> op[100100];
  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 >> l[i] >> r[i] >> d[i];
  }
  int x, y;
  for (int i = 1; i <= k; i++) {
    cin >> x >> y;
    op[x].first++;
    op[y].second++;
  }
  long long int xx = 0;
  for (int i = 1; i <= m; i++) {
    xx += op[i].first;
    oper[i] = xx;
    xx -= op[i].second;
  }
  xx = 0;
  for (int i = 1; i <= m; i++) {
    p[l[i]].first += (d[i] * oper[i]);
    p[r[i]].second += (d[i] * oper[i]);
  }
  for (int i = 1; i <= n; i++) {
    xx += p[i].first;
    b[i] = xx;
    b[i] += a[i];
    cout << b[i] << " ";
    xx -= p[i].second;
  }
}

```