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;
int n, mx = 0, a[200007], b[200007], q;
vector<pair<int, pair<int, int>>> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n;
  for (long long i = 1; i <= (long long)n; ++i) cin >> a[i];
  cin >> q;
  for (long long i = 1; i <= (long long)q; ++i) {
    int h, x, y;
    cin >> h;
    if (h == 1) {
      cin >> x >> y;
      v.push_back({1, {x, y}});
    } else {
      cin >> x;
      v.push_back({2, {x, -1}});
    }
  }
  reverse(((v).begin()), ((v).end()));
  int mx = 0;
  for (long long i = 0; i <= (long long)q - 1; ++i) {
    pair<int, pair<int, int>> p = v[i];
    if (p.first == 1) {
      if (b[p.second.first]) continue;
      a[p.second.first] = max(mx, p.second.second);
      b[p.second.first] = 1;
    } else
      mx = max(mx, p.second.first);
  }
  for (long long i = 1; i <= (long long)n; ++i) {
    if (!b[i]) a[i] = max(a[i], mx);
    cout << a[i] << " ";
  }
}
