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
input=sys.stdin.readline
def construct(array,cur_pos,tree,start,end):
	if(start==end):
		tree[cur_pos]=array[start]
	else:
		mid=(start+end)//2
		tree[cur_pos]=max(construct(array,2*cur_pos+1,tree,start,mid),construct(array,2*cur_pos+2,tree,mid+1,end))
	return tree[cur_pos]
def query(start,end,l,r,index,tree):
	if(l<=start and r>=end):
		return tree[index]
	if(end<l or start>r):
		return -1
	mid=(start+end)//2
	return max(query(start,mid,l,r,2*index+1,tree),query(mid+1,end,l,r,2*index+2,tree))
n=int(input())
l= list(map(int,input().split()))
q=int(input())
last_query=[-1]*n
queries=[-1]*q
for i in range(q):
 
	l1= list(map(int,input().split()))
	if(l1[0]==1):
		index=l1[1]-1
		value=l1[2]
		l[index]=value
		last_query[index]=i
	else:
		queries[i]=l1[1]
	
tree=[0]*(5*q)
array=queries
construct(array,0,tree,0,q-1)
 
for i in range(n):
	temp=last_query[i]
	if(temp+1<q):
		q1=query(0,q-1,temp+1,q-1,0,tree)
 
		l[i]=max(q1,l[i])

print(*l, sep = " ")  
# print(s)
# print("--- %s seconds ---" % (time.time() - start_time))
