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;
const long long imx = INT_MAX;
const long long imn = INT_MIN;
int static dp[1001][1001];
bool comp(pair<long long, long long> &a, pair<long long, long long> &b) {
  return a.second < b.second;
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long(i) = 0; (i) < (n); (i)++) {
    cin >> a[i];
  }
  long long q;
  cin >> q;
  vector<pair<long long, long long>> v(n);
  for (long long(i) = 0; (i) < (n); (i)++) {
    v[i] = {-1, a[i]};
  }
  vector<pair<long long, long long>> d(q);
  for (long long(i) = 0; (i) < (q); (i)++) {
    d[i] = {i, 0};
    long long t;
    cin >> t;
    if (t == 1) {
      long long p, x;
      cin >> p >> x;
      v[p - 1] = {i, x};
    } else {
      long long x;
      cin >> x;
      d[i] = {i, x};
    }
  }
  for (long long(i) = (q - 2); (i) >= (0); (i)--) {
    d[i].second = max(d[i].second, d[i + 1].second);
  }
  sort(d.begin(), d.end());
  for (long long(i) = 0; (i) < (n); (i)++) {
    if (v[i].first == -1) {
      a[i] = max(d[0].second, v[i].second);
    } else {
      auto x = lower_bound(d.begin(), d.end(), v[i]);
      if (x == d.end()) {
        a[i] = v[i].second;
      } else {
        a[i] = max((*x).second, v[i].second);
      }
    }
  }
  for (long long(i) = 0; (i) < (n); (i)++) {
    cout << a[i] << " ";
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (long long(_) = 0; (_) < (t); (_)++) {
    solve();
  }
  return 0;
}
