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;
void Main();
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  while (~cin.peek()) Main(), cin.get();
  return 0;
}
void Main() {
  int n;
  cin >> n;
  vector<int> a(n + 1);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  vector<int> count(n + 1);
  int m;
  cin >> m;
  vector<int> maxv;
  for (int i = 1; i <= m; i++) {
    int t, p, x;
    cin >> t;
    if (t == 1) {
      cin >> p >> x;
      a[p] = x;
      count[p] = maxv.size();
    } else {
      cin >> x;
      maxv.push_back(x);
    }
  }
  maxv.push_back(0);
  for (int i = maxv.size() - 2; i >= 0; i--) {
    maxv[i] = max(maxv[i], maxv[i + 1]);
  }
  for (int i = 1; i <= n; i++) {
    cout << max(a[i], maxv[count[i]]) << " \n"[i == n];
  }
}
