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 maxn = 4e5 + 5;
pair<int, pair<int, int>> q[maxn];
int n, qunm;
bool vis[maxn];
int a[maxn];
signed main() {
  cin >> n;
  for (int i = 0; i < n; i++)
    cin >> q[i].second.second, q[i].first = 1, q[i].second.first = i + 1;
  cin >> qunm;
  for (int i = n; i < n + qunm; i++) {
    int x;
    cin >> x;
    if (x == 1) {
      cin >> q[i].second.first >> q[i].second.second;
      q[i].first = 1;
    } else {
      cin >> q[i].second.first;
      q[i].first = 2;
    }
  }
  int mx = -1;
  for (int i = n + qunm - 1; i >= 0; i--) {
    if (q[i].first == 2)
      mx = max(mx, q[i].second.first);
    else {
      if (vis[q[i].second.first])
        continue;
      else {
        a[q[i].second.first] = max(mx, q[i].second.second);
        vis[q[i].second.first] = 1;
      }
    }
  }
  for (int i = 1; i <= n; i++) cout << a[i] << ' ';
}
