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 mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-9;
const int MAX_SIZE = 50;
const int dX[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int dY[] = {0, 0, -1, 1, 1, -1, -1, 1};
template <class S, class T>
ostream& operator<<(ostream& o, const pair<S, T>& a) {
  o << a.first << " " << a.second << " ";
  return o;
}
template <class S, class T>
ostream& operator<<(ostream& o, const map<S, T>& a) {
  for (auto& it : (a)) o << (it);
  return o;
}
template <class S>
ostream& operator<<(ostream& o, const vector<S>& a) {
  for (auto& it : (a)) o << (it) << " ";
  o << "\n";
  return o;
}
template <class S>
ostream& operator<<(ostream& o, const vector<vector<S> >& a) {
  for (auto& it : (a)) o << (it);
  return o;
}
template <class S>
ostream& operator<<(ostream& o, const set<S>& a) {
  for (auto& it : (a)) o << (it) << " ";
  o << "\n";
  return o;
}
const long long MAX = 100010;
long long a[MAX], c[MAX], cnt[MAX], l[MAX], r[MAX], d[MAX];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int i, n, m, k, x, y;
  long long val;
  cin >> n >> m >> k;
  for (i = (0); (((1) > 0) - ((1) < 0)) * i < (((1) > 0) - ((1) < 0)) * (n);
       i += (1))
    cin >> a[i];
  for (i = (0); (((1) > 0) - ((1) < 0)) * i < (((1) > 0) - ((1) < 0)) * (m);
       i += (1)) {
    cin >> l[i] >> r[i] >> d[i];
    l[i]--;
    r[i]--;
  }
  for (i = (0); (((1) > 0) - ((1) < 0)) * i < (((1) > 0) - ((1) < 0)) * (k);
       i += (1)) {
    cin >> x >> y;
    x--;
    cnt[x]++;
    cnt[y]--;
  }
  for (i = (1); (((1) > 0) - ((1) < 0)) * i < (((1) > 0) - ((1) < 0)) * (m);
       i += (1))
    cnt[i] += cnt[i - 1];
  for (i = (0); (((1) > 0) - ((1) < 0)) * i < (((1) > 0) - ((1) < 0)) * (m);
       i += (1)) {
    val = cnt[i] * d[i];
    c[l[i]] += val;
    c[r[i] + 1] -= val;
  }
  for (i = (1); (((1) > 0) - ((1) < 0)) * i < (((1) > 0) - ((1) < 0)) * (n);
       i += (1))
    c[i] += c[i - 1];
  for (i = (0); (((1) > 0) - ((1) < 0)) * i < (((1) > 0) - ((1) < 0)) * (n);
       i += (1)) {
    cout << (a[i] + c[i]) << " ";
  }
  return 0;
}
