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 long long MAX = 1e5 + 1;
long long n, m, k, arr[MAX];
pair<long long, pair<long long, long long>> op[MAX];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  arr[0] = 0;
  for (long long i = 1; i <= m; i++) {
    long long l, r, d;
    cin >> l >> r >> d;
    op[i] = make_pair(d, make_pair(l, r));
  }
  vector<long long> freq(m + 1, 0);
  vector<long long> dp(m + 2, 0);
  for (long long i = 1; i <= k; i++) {
    long long l, r;
    cin >> l >> r;
    dp[l] = dp[l] + 1;
    dp[r + 1] = dp[r + 1] - 1;
  }
  long long sum = 0;
  for (long long i = 1; i <= m; i++) {
    sum += dp[i];
    freq[i] = sum;
  }
  vector<long long> dp2(n + 2, 0);
  for (long long i = 1; i <= n; i++) {
    dp2[i] = arr[i] - arr[i - 1];
  }
  for (long long i = 1; i <= m; i++) {
    long long val = op[i].first;
    long long l = op[i].second.first;
    long long r = op[i].second.second;
    dp2[l] = dp2[l] + val * freq[i];
    dp2[r + 1] = dp2[r + 1] - val * freq[i];
  }
  sum = 0;
  for (long long i = 1; i <= n; i++) {
    sum += dp2[i];
    arr[i] = sum;
    cout << arr[i] << " ";
  }
  cout << endl;
  return 0;
}
