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
import sys

in_file = sys.stdin #open("D.txt", "r")

n = int(in_file.readline().strip())
A = list(map(int, in_file.readline().strip().split()))
q = int(in_file.readline().strip())

flood = []
last = [0 for a in A]
for i in range(q):
    t = list(map(int, in_file.readline().strip().split()))
    if t[0] == 1:
        A[t[1]-1] = t[2]
        last[t[1]-1] = i
        flood.append(0)
    else:
        flood.append(t[1])
flood_max = [flood[-1]]
for i in range(len(flood)-1):
    flood_max.append(max(flood_max[-1],flood[len(flood)-i-2]))
flood_max.reverse()
for i in range(len(A)):
    A[i] = max(A[i], flood_max[last[i]])
sys.stdout.write(" ".join(map(str, A)))
sys.stdout.flush()