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 ps[200100];
int N;
int nowmoney[200100];
vector<int> noplist;
vector<int> payout;
int maxcals[200100];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> N;
  int i;
  for (i = 0; i < N; i++) {
    cin >> nowmoney[i];
  }
  int M;
  cin >> M;
  for (i = 0; i < M; i++) {
    int a, b, c;
    cin >> a;
    if (a == 1) {
      cin >> b >> c;
      noplist.push_back(b - 1);
      nowmoney[b - 1] = c;
      ps[b - 1] = -1;
    } else {
      cin >> b;
      int j;
      for (j = 0; j < noplist.size(); j++) {
        ps[noplist[j]] = payout.size();
      }
      payout.push_back(b);
      noplist.clear();
    }
  }
  for (i = payout.size() - 1; i >= 0; i--) {
    maxcals[i] = max(maxcals[i + 1], payout[i]);
  }
  for (i = 0; i < N; i++) {
    if (ps[i] >= 0)
      cout << max(nowmoney[i], maxcals[ps[i]]) << ' ';
    else
      cout << nowmoney[i] << ' ';
  }
}
