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 int N = 2e5 + 10;
int a[N], ma[N];
bool vis[N];
struct str {
  int id, va;
} b[N];
int main() {
  int n, q;
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  while (cin >> n) {
    for (int i = 1; i <= n; i++) cin >> a[i];
    memset(vis, 0, sizeof(vis));
    memset(ma, -1, sizeof(ma));
    cin >> q;
    int op, p, x;
    for (int i = 1; i <= q; i++) {
      cin >> op;
      if (op == 1) {
        cin >> p >> x;
        vis[p] = 1;
        b[p].id = i;
        b[p].va = x;
      } else {
        cin >> x;
        ma[i] = x;
      }
    }
    for (int i = q; i > 0; i--) {
      ma[i] = max(ma[i + 1], ma[i]);
    }
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) {
        if (ma[1] > a[i])
          cout << ma[1] << " ";
        else
          cout << a[i] << " ";
      } else {
        int no = b[i].id;
        a[i] = b[i].va;
        if (ma[no] > a[i])
          cout << ma[no] << " ";
        else
          cout << a[i] << " ";
      }
    }
    cout << endl;
  }
  return 0;
}
