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
def main():
    n = int(input())
    money = list(map(int,input().split()))
    last_change = [-1]*n
    dp = []
    q = int(input())
    second = []
    for i in range(q):
        query = list(map(int,input().split()))
        if query[0] == 1:
            dp.append(-1)
            p,x = query[1],query[2]
            money[p-1] = x
            last_change[p-1] = i
        else:
            second.append(query[1])
            dp.append(query[1])

    for i in range(len(dp)-2,-1,-1):
        dp[i] = max(dp[i+1],dp[i])


    for i in range(n):
        if last_change[i] == -1:
            money[i] = max(dp[0],money[i])
        else:
            change = last_change[i]
            if change+1 < len(dp):
                max_val = dp[change+1]
            else:
                max_val = -1
            money[i] = max(max_val,money[i])

    for i in money:
        print(i,end = ' ')
    


main()
