There is a country with n citizens. The i-th of them initially has a_{i} money. The government strictly controls the wealth of its citizens. Whenever a citizen makes a purchase or earns some money, they must send a receipt to the social services mentioning the amount of money they currently have.

Sometimes the government makes payouts to the poor: all citizens who have strictly less money than x are paid accordingly so that after the payout they have exactly x money. In this case the citizens don't send a receipt.

You know the initial wealth of every citizen and the log of all events: receipts and payouts. Restore the amount of money each citizen has after all events.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the numer of citizens.

The next line contains n integers a_1, a_2, ..., a_n (0 ≤ a_{i} ≤ 10^{9}) — the initial balances of citizens.

The next line contains a single integer q (1 ≤ q ≤ 2 ⋅ 10^{5}) — the number of events.

Each of the next q lines contains a single event. The events are given in chronological order.

Each event is described as either 1 p x (1 ≤ p ≤ n, 0 ≤ x ≤ 10^{9}), or 2 x (0 ≤ x ≤ 10^{9}). In the first case we have a receipt that the balance of the p-th person becomes equal to x. In the second case we have a payoff with parameter x.

Output

Print n integers — the balances of all citizens after all events.

Examples

Input


4
1 2 3 4
3
2 3
1 2 2
2 1


Output


3 2 3 4 


Input


5
3 50 2 1 10
3
1 2 0
2 8
1 3 20


Output


8 8 20 8 10 

Note

In the first example the balances change as follows: 1 2 3 4 → 3 3 3 4 → 3 2 3 4 → 3 2 3 4

In the second example the balances change as follows: 3 50 2 1 10 → 3 0 2 1 10 → 8 8 8 8 10 → 8 8 20 8 10
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
int a[N];
pair<int, int> lst[N];
vector<pair<int, int> > sec;
deque<int> ps;
int res[N];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], lst[i] = {0, a[i]};
  int q;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      int p, x;
      cin >> p >> x;
      lst[p] = {i, x};
    } else {
      int x;
      cin >> x;
      sec.push_back({i, x});
    }
  }
  int mx = 0;
  for (int i = sec.size() - 1; i >= 0; i--) {
    mx = max(mx, sec[i].second);
    ps.push_front(mx);
  }
  for (int i = 1; i <= n; i++) {
    int tofind = lst[i].first;
    int r = sec.size(), l = -1, m;
    while (r - l > 1) {
      m = (l + r) / 2;
      if (sec[m].first <= tofind)
        l = m;
      else
        r = m;
    }
    if (r == sec.size())
      res[i] = lst[i].second;
    else
      res[i] = max(lst[i].second, ps[r]);
  }
  for (int i = 1; i <= n; i++) cout << res[i] << " ";
  return 0;
}
