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;
struct qtype {
  int flag, p, x;
};
int main() {
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  int q;
  cin >> q;
  qtype b[q];
  for (int i = 0; i < q; i++) {
    int flag, p = -1, x;
    cin >> flag;
    if (flag == 1) {
      cin >> p;
      p--;
    }
    cin >> x;
    b[i] = {flag, p, x};
  }
  int mx2 = -1;
  bool mark[n];
  for (int i = 0; i < n; i++) mark[i] = false;
  for (int i = q - 1; i >= 0; i--) {
    if (b[i].flag == 1) {
      if (mark[b[i].p]) continue;
      a[b[i].p] = b[i].x;
      a[b[i].p] = max(a[b[i].p], mx2);
      mark[b[i].p] = true;
    } else {
      mx2 = max(mx2, b[i].x);
    }
  }
  for (int i = 0; i < n; i++)
    if (!mark[i]) a[i] = max(a[i], mx2);
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  puts("");
}
