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 int MAXN = 2e5 + 1;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  int q;
  int a[MAXN];
  cin >> n;
  for (int i = 0; i < n; i += 1) cin >> a[i];
  cin >> q;
  int c = 0;
  vector<int> v;
  int lstv[MAXN];
  int lsti[MAXN];
  memset(lstv, -1, sizeof(lstv));
  ;
  int mx[MAXN];
  for (int i = 0; i < q; i += 1) {
    int p, x;
    cin >> c;
    if (c == 1) {
      cin >> p >> x;
      p--;
      lstv[p] = x;
      lsti[p] = i;
      v.push_back(-1);
    }
    if (c == 2) {
      cin >> x;
      v.push_back(x);
    }
  }
  int max_ = -1;
  for (int i = v.size() - 1; i >= 0; i--) {
    max_ = max({max_, v[i]});
    mx[i] = max_;
  }
  for (int i = 0; i < n; i += 1) {
    if (lstv[i] != -1)
      a[i] = max({lstv[i], mx[lsti[i]]});
    else
      a[i] = max({a[i], mx[0]});
  }
  for (int i = 0; i < n; i += 1) cout << a[i] << " ";
  cout << endl;
  return 0;
}
