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;
int main() {
  long long n, m, k;
  cin >> n >> m >> k;
  vector<unsigned long long> a(n);
  vector<long long> l(m + 1);
  vector<int> r(m + 1);
  vector<unsigned long long> p(m + 1);
  unordered_map<long long, long long> d;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < m; i++) {
    cin >> l[i] >> r[i] >> p[i];
  }
  int x, y;
  vector<unsigned long long> count(m + 1, 0);
  vector<unsigned long long> frero(n + 1, 0);
  for (int i = 0; i < k; i++) {
    cin >> x >> y;
    count[x - 1]++;
    count[y]--;
  }
  for (int i = 0; i < m; i++) {
    if (i) count[i] += count[i - 1];
    p[i] *= count[i];
    frero[l[i] - 1] += p[i];
    frero[r[i]] -= p[i];
  }
  for (int i = 0; i < n; i++) {
    if (i) frero[i] += frero[i - 1];
    a[i] += frero[i];
    cout << a[i] << " ";
  }
}
