Welfare State

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 $n$ ($1 \le n \le 2 \cdot 10^{5}$) — the numer of citizens.

The next line contains $n$ integers $a_1$, $a_2$, …, $a_n$ ($0 \le a_{i} \le 10^{9}$) — the initial balances of citizens.

The next line contains a single integer $q$ ($1 \le q \le 2 \cdot 10^{5}$) — 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 ($1 \le p \le n$, $0 \le x \le 10^{9}$), or 2 x ($0 \le x \le 10^{9}$). 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 $\rightarrow$ 3 3 3 4 $\rightarrow$ 3 2 3 4 $\rightarrow$ 3 2 3 4

In the second example the balances change as follows: 3 50 2 1 10 $\rightarrow$ 3 0 2 1 10 $\rightarrow$ 8 8 8 8 10 $\rightarrow$ 8 8 20 8 10

Solution:

#include <bits/stdc++.h>

using namespace std;

int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) {
    cin >> a[i];
}
int tt;
cin >> tt;
vector<int> op(tt);
vector<int> who(tt);
vector<int> val(tt);
for (int i = 0; i < tt; i++) {
    cin >> op[i];
if (op[i] == 1) {
cin >> who[i] >> val[i];
--who[i];
} else {
cin >> val[i];
}
}
vector<int> res(n, -1);
int mx = -1;
for (int i = tt - 1; i >= 0; i--) {
if (op[i] == 1) {
if (res[who[i]] == -1) {
res[who[i]] = max(val[i], mx);
}
} else {
mx = max(mx, val[i]);
}
}
for (int i = 0; i < n; i++) {
    if (res[i] == -1) {
      res[i] = max(a[i], mx);
    }
  }
  for (int i = 0; i < n; i++) {
    if (i > 0) {
cout << " ";
    }
    cout << res[i];
  }
  cout << '\n';
  return 0;
}