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 mod = 1e9 + 7;
struct cmp {
  bool operator()(const pair<long long, long long> &a,
                  const pair<long long, long long> &b) const {
    return a.first - a.second + 1 > b.first - b.second + 1;
  }
};
void solve() {
  long long n;
  cin >> n;
  vector<long long> arr(n);
  for (auto &u : arr) cin >> u;
  long long q;
  cin >> q;
  vector<long long> typ(q), val(q), ind(q, -1), what(q, 0);
  set<long long> done;
  for (long long i = 0; i < q; i++) {
    cin >> typ[i];
    if (typ[i] == 1) {
      cin >> ind[i] >> val[i];
      ind[i]--;
      done.insert(ind[i]);
    } else {
      cin >> what[i];
    }
  }
  for (long long i = q - 1; i >= 0; i--) {
    if (i == q - 1) continue;
    what[i] = max(what[i], what[i + 1]);
  }
  for (long long i = 0; i < q; i++) {
    if (typ[i] == 1) {
      arr[ind[i]] = max(val[i], what[i]);
    }
  }
  for (long long i = 0; i < n; i++) {
    if (done.find(i) == done.end())
      cout << max(arr[i], what[0]) << " ";
    else
      cout << arr[i] << " ";
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long test = 1;
  while (test--) {
    solve();
  }
  return 0;
}
