#include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long long N = 3e5 + 5;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const int INFi = 0x7f7f7f7f;
long long test = 1, n, m, k, a[N], pre[N], pref[N], op[N][3];
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> op[i][0] >> op[i][1] >> op[i][2];
  while (k--) {
    int x, y;
    cin >> x >> y;
    pre[x]++, pre[y + 1]--;
  }
  for (int i = 1; i <= m; i++) {
    pre[i] += pre[i - 1];
    pref[op[i][0]] += pre[i] * op[i][2];
    pref[op[i][1] + 1] -= pre[i] * op[i][2];
  }
  for (int i = 1; i <= n; i++) {
    pref[i] += pref[i - 1];
    cout << pref[i] + a[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  for (int i = 1; i <= test; i++) {
    solve();
  }
}

### 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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const long long N = 3e5 + 5;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
const int INFi = 0x7f7f7f7f;
long long test = 1, n, m, k, a[N], pre[N], pref[N], op[N][3];
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= m; i++) cin >> op[i][0] >> op[i][1] >> op[i][2];
  while (k--) {
    int x, y;
    cin >> x >> y;
    pre[x]++, pre[y + 1]--;
  }
  for (int i = 1; i <= m; i++) {
    pre[i] += pre[i - 1];
    pref[op[i][0]] += pre[i] * op[i][2];
    pref[op[i][1] + 1] -= pre[i] * op[i][2];
  }
  for (int i = 1; i <= n; i++) {
    pref[i] += pref[i - 1];
    cout << pref[i] + a[i] << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  for (int i = 1; i <= test; i++) {
    solve();
  }
}

```