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 long long int mod = 1e9 + 7;
long long int power(long long int x, long long int y) {
  if (y == 0) return 1;
  long long int p = power(x, y / 2) % mod;
  p = (p * p) % mod;
  return (y % 2 == 0) ? p : (x * p) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  long long int a[n + 1], b[n + 1];
  for (long long int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long int q;
  cin >> q;
  long long int c[q + 1];
  memset(c, 0, sizeof(c));
  memset(b, 0, sizeof(b));
  for (long long int qq = 1; qq <= q; qq++) {
    long long int type;
    cin >> type;
    if (type == 1) {
      long long int p, x;
      cin >> p >> x;
      a[p] = x;
      b[p] = qq;
    } else if (type == 2) {
      long long int x;
      cin >> x;
      c[qq] = x;
    }
  }
  for (long long int i = q - 1; i >= 1; i--) {
    c[i] = max(c[i], c[i + 1]);
  }
  c[0] = c[1];
  for (long long int i = 1; i <= n; i++) {
    a[i] = max(a[i], c[b[i]]);
  }
  for (long long int i = 1; i <= n; i++) cout << a[i] << " ";
}
