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;
struct query {
  long long int type, a, b;
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    long long int arr[n];
    for (long long int i = 0; i < n; i++) cin >> arr[i];
    long long int q;
    cin >> q;
    vector<query> que;
    for (long long int i = 0; i < q; i++) {
      query h;
      cin >> h.type;
      if (h.type == 1)
        cin >> h.a >> h.b;
      else {
        cin >> h.a;
        h.b = 0;
      }
      que.push_back(h);
    }
    reverse(que.begin(), que.end());
    long long int maxi = INT_MIN;
    bool visited[n];
    memset(visited, false, sizeof(visited));
    for (long long int i = 0; i < q; i++) {
      if (que[i].type == 1 && !visited[que[i].a - 1]) {
        arr[que[i].a - 1] = max(que[i].b, maxi);
        visited[que[i].a - 1] = true;
      } else if (que[i].type == 2)
        maxi = max(maxi, que[i].a);
    }
    for (long long int i = 0; i < n; i++) {
      if (arr[i] < maxi && !visited[i]) arr[i] = maxi;
      cout << arr[i] << " ";
    }
  }
  return 0;
}
