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 used[200001];
int main() {
  long long n;
  cin >> n;
  int a[n + 1];
  for (int k = 1; k <= n; k++) cin >> a[k];
  long long m;
  cin >> m;
  pair<long long, long long> cnt[m + 1];
  for (int k = 1; k <= m; k++) {
    int x;
    cin >> x;
    if (x == 1) {
      int y, z;
      cin >> y >> z;
      cnt[k].first = y;
      cnt[k].second = z;
    }
    if (x == 2) {
      int y;
      cin >> y;
      cnt[k].first = y;
      cnt[k].second = -1;
    }
  }
  long long mx = -1;
  long long ans[n + 1];
  for (int k = m; k > 0; k--) {
    if (cnt[k].second == -1) {
      mx = max(mx, cnt[k].first);
    }
    if (cnt[k].second != -1 && used[cnt[k].first] == 0) {
      ans[cnt[k].first] = max(mx, cnt[k].second);
      used[cnt[k].first] = 1;
    }
  }
  for (int k = 1; k <= n; k++) {
    if (used[k] == 0) {
      long long xx = a[k];
      cout << max(xx, mx) << ' ';
    } else
      cout << ans[k] << ' ';
  }
  return 0;
}
