1199_D. 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 ≤ 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
```python
import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

def main():

    n = int(input())
    A = list(map(int, input().split()))
    q = int(input())
    Q = []
    for i in range(q):
        temp = list(map(int, input().split()))
        t = temp[0]
        if t == 1:
            p, x = temp[1], temp[2]
            p -= 1
            Q.append((t, p, x))
        else:
            x = temp[1]
            Q.append((t, -1, x))
    Q.reverse()
    B = [-1]*n
    y = -1
    for t, p, x in Q:
        if t == 1:
            if B[p] == -1:
                B[p] = max(y, x)
        else:
            y = max(y, x)
    #print(*B)
    for i, b in enumerate(B):
        if b != -1:
            continue
        B[i] = max(A[i], y)
    print(*B)

if __name__ == '__main__':
    main()

```