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 $n$ ($1 \le n \le 2 \cdot 10^{5}$) — the numer of citizens.

The next line contains $n$ integers $a_1$, $a_2$, ..., $a_n$ ($0 \le a_{i} \le 10^{9}$) — the initial balances of citizens.

The next line contains a single integer $q$ ($1 \le q \le 2 \cdot 10^{5}$) — 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 ($1 \le p \le n$, $0 \le x \le 10^{9}$), or 2 x ($0 \le x \le 10^{9}$). 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 $\rightarrow$ 3 3 3 4 $\rightarrow$ 3 2 3 4 $\rightarrow$ 3 2 3 4

In the second example the balances change as follows: 3 50 2 1 10 $\rightarrow$ 3 0 2 1 10 $\rightarrow$ 8 8 8 8 10 $\rightarrow$ 8 8 20 8 10

```python

USER_FILENAME = "<user_code>"
import sys, json, builtins
import math, heapq, bisect, itertools, functools, collections, statistics, fractions, decimal, operator, array, random, re, string, os, pathlib
try:
    import resource as _res
except Exception:
    _res = None
from collections import Counter, defaultdict, deque, OrderedDict
from functools import lru_cache, reduce
from itertools import accumulate, permutations, combinations, product
from math import gcd, ceil, floor, sqrt
from decimal import Decimal
try:
    from typing import *
except Exception:
    pass
try:
    sys.setrecursionlimit(1_000_000)
except Exception:
    pass
try:
    import site as _site
    try:
        _sys_sites = _site.getsitepackages() if hasattr(_site, "getsitepackages") else []
        for _p in _sys_sites:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
    try:
        _usp = _site.getusersitepackages()
        if isinstance(_usp, str):
            _usp = [_usp]
        for _p in _usp:
            if _p and _p not in sys.path:
                sys.path.append(_p)
    except Exception:
        pass
except Exception:
    pass

ns = {"__name__": "__main__"}
for _k, _v in list(globals().items()):
    if not _k.startswith("_"):
        ns[_k] = _v
_CODE = 'n = int(input())\nbalance = list(map(int, input().split()))\nq = int(input())\nlog = [list(map(int, input().split())) for i in range(q)]\n\nind_of_last_operation = [None] * n\nmax_pay = 0\nind_of_max_payment = None\npayments = []\npayments_id = []\n\nfor i in range(q):\n    if log[i][0] == 1:\n        ind_of_last_operation[log[i][1] - 1] = i\n\nfor i in range(q):\n    if log[i][0] == 2:\n        ind_of_last_payment = i\n        last_payment = log[i][1]\n        if log[i][1] > max_pay:\n            max_pay = log[i][1]\n\nmax_pay_for_moment = [0] * (q + 1)\nfor j in range(q - 1, -1, -1):\n    if log[j][0] == 2:\n        max_pay_for_moment[j] = max(log[j][1], max_pay_for_moment[j + 1])\n    else:\n        max_pay_for_moment[j] = max_pay_for_moment[j + 1]\n\nfor i in range(n):\n    if ind_of_last_operation[i] == None:\n        if balance[i] < max_pay:\n            balance[i] = max_pay\n    else:\n        if max_pay_for_moment[ind_of_last_operation[i] + 1] > log[ind_of_last_operation[i]][2]:\n            balance[i] = max_pay_for_moment[ind_of_last_operation[i] + 1]\n        else:\n            balance[i] = log[ind_of_last_operation[i]][2]\n\nprint(*balance)'
_co = compile(_CODE, USER_FILENAME, "exec")
exec(_co, ns, ns)

```

Measured memory_bytes: 13693