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 bisect,sys
printn = lambda x: sys.stdout.write(x)
inn = lambda : int(input())
inl   = lambda: list(map(int, input().split()))
inm   = lambda:      map(int, input().split())
DBG = True  and False
R = 10**9 + 7

def ddprint(x):
  if DBG:
    print(x)

n = inn()
a = inl()
q = inn()
spend = []
pay = []
mx = 0
for i in range(q):
    c = inl()
    if c[0]==1:
        spend.append((i,c[1],c[2]))
    else:
        pay.append((i,c[1]))
        if c[1]>mx:
            mx = c[1]

m = len(pay)
accpay = [(0,0)] * (m+1)
for i in range(m-1,-1,-1):
    accpay[i] = (pay[i][0], max(accpay[i+1][1], pay[i][1]))
accpay[m] = (999999999,0)

b = [0]*n
for i in range(n):
    b[i] = max(a[i],mx)
for z in spend:
    idx = bisect.bisect_left(accpay, (z[0],z[1]))
    if idx >= m:
        b[z[1]-1] = z[2]
    else:
        b[z[1]-1] = max(z[2], accpay[idx][1])

for i in range(n):
    printn(("" if i==0 else " ") + str(b[i]))
print("")
