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;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long n, i, j, k, q, type, p, x, cnt = 0, mx = -1, op, query;
  cin >> n;
  vector<long long> v(n);
  for (i = 0; i < n; i++) {
    cin >> v[i];
  }
  cin >> q;
  query = q;
  map<long long, long long> mp, mpp;
  while (q--) {
    cin >> type;
    cnt++;
    if (type == 2) {
      cin >> x;
      mp[cnt] = x;
      mx = max(mx, x);
    } else {
      cin >> p >> x;
      v[p - 1] = x;
      mpp[p - 1] = cnt;
    }
  }
  for (i = query - 1; i >= 1; i--) mp[i] = max(mp[i], mp[i + 1]);
  for (i = 0; i < n; i++) {
    if (mpp[i]) {
      if (v[i] < mp[mpp[i]]) v[i] = mp[mpp[i]];
    } else if (v[i] < mx)
      v[i] = mx;
  }
  for (i = 0; i < n; i++) cout << v[i] << " ";
  cout << endl;
}
