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>
const long long M = 998244353;
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  long long a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  long long x = -1, fin = -1;
  map<long long, long long> mp;
  vector<long long> v;
  set<long long> s;
  long long q;
  cin >> q;
  while (q--) {
    long long m;
    cin >> m;
    long long p, y;
    if (m == 1) {
      cin >> p >> y;
      mp[p - 1] = v.size();
      s.insert(p - 1);
      a[p - 1] = y;
    } else {
      cin >> y;
      v.push_back(y);
    }
  }
  if (v.size() == 0) {
    for (int i = 0; i < n; i++) cout << a[i] << " ";
    return 0;
  }
  vector<long long> max1(v.size());
  max1[v.size() - 1] = v.back();
  for (int i = v.size() - 2; i >= 0; i--) {
    if (v[i] > max1[i + 1])
      max1[i] = v[i];
    else
      max1[i] = max1[i + 1];
  }
  map<long long, long long>::iterator it;
  for (it = mp.begin(); it != mp.end(); it++) {
    if (it->second == 0) {
      a[it->first] = max(a[it->first], max1[0]);
    } else {
      if (it->second >= v.size())
        continue;
      else {
        a[it->first] = max(a[it->first], max1[it->second]);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    if (mp.find(i) == mp.end()) {
      a[i] = max(a[i], max1[0]);
    }
  }
  for (int i = 0; i < n; i++) cout << a[i] << " ";
  return 0;
}
