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;
long long cnum[400005] = {0};
int main() {
  long long n;
  cin >> n;
  long long a[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long q, x, y, z, cm = 0;
  cin >> q;
  std::vector<pair<long long, pair<long long, long long> > > p;
  std::vector<long long> q2;
  for (int i = 0; i < q; i++) {
    cin >> x;
    if (x == 1) {
      cin >> y >> z;
      p.push_back({x, {y, z}});
    } else {
      cm++;
      cin >> y;
      p.push_back({x, {y, -1}});
      q2.push_back(y);
    }
  }
  long long maxi = -1, dontexecq2[cm];
  for (int i = cm - 1; i >= 0; i--) {
    if (q2[i] > maxi) {
      dontexecq2[i] = 1;
      maxi = max(maxi, q2[i]);
    } else {
      dontexecq2[i] = 0;
    }
  }
  long long cmcount = 0, flag = 0;
  set<long long> s;
  for (int i = 0; i < q; i++) {
    if (p[i].first == 1) {
      a[p[i].second.first - 1] = p[i].second.second;
      if (flag == 1) {
        s.insert(p[i].second.first - 1);
      }
    } else {
      if (dontexecq2[cmcount] == 0) {
        cmcount++;
        continue;
      }
      z = p[i].second.first;
      if (flag == 0) {
        flag = 1;
        s.clear();
        for (int j = 0; j < n; j++) {
          a[j] = max(a[j], z);
        }
      } else {
        set<long long>::iterator itr = s.begin();
        for (itr = s.begin(); itr != s.end(); itr++) {
          a[*itr] = max(a[*itr], z);
        }
        s.clear();
      }
      cmcount++;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << a[i] << ' ';
  }
}
