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;
pair<int, int> a[200007];
int b[200007];
int res[200007];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int i = 1;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int b;
    cin >> b;
    a[i] = {0, b};
  }
  int q;
  cin >> q;
  int z = q;
  while (q--) {
    int p;
    cin >> p;
    int l, v;
    if (p == 1) {
      cin >> l >> v;
      a[l - 1] = {i, v};
      i++;
    } else {
      cin >> v;
      b[i - 1] = max(b[i - 1], v);
    }
  }
  for (int k = z; k >= 0; k--) {
    b[k] = max(b[k + 1], b[k]);
  }
  for (int k = 0; k < n; k++) {
    res[k] = max(a[k].second, b[a[k].first]);
    cout << res[k] << " ";
  }
  cout << ("") << '\n';
  return 0;
}
