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;
long long int exp(long long int a, long long int b, long long int m);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) cin >> a[i];
  long long int q;
  cin >> q;
  long long int n1 = n;
  n = q;
  long long int ind = -1;
  long long int aq[q][3];
  for (long long int i = 0; i < n; i++) {
    cin >> aq[i][0];
    cin >> aq[i][1];
    if (aq[i][0] == 1) {
      cin >> aq[i][2];
    }
  }
  long long int su[n];
  long long int x = 0;
  for (long long int i = 0; i < n; i++) {
    if (aq[n - i - 1][0] == 2) x = max(x, aq[n - i - 1][1]);
    su[n - i - 1] = x;
  }
  for (long long int i = 0; i < n1; i++) a[i] = max(a[i], x);
  for (long long int i = 0; i < n; i++) {
    if (aq[i][0] == 1) {
      a[aq[i][1] - 1] = max(su[i], aq[i][2]);
    }
  }
  for (long long int i = 0; i < n1; i++) cout << a[i] << " ";
  cout << endl;
  return 0;
}
long long int exp(long long int a, long long int b, long long int m) {
  if (b == 0) {
    return 1;
  }
  long long int temp = exp(a, b / 2, m);
  temp = (temp * temp) % m;
  if (b & 1) {
    return (temp * (a % m)) % m;
  }
  return temp;
}
