Ridbit starts with an integer n.

In one move, he can perform one of the following operations: 

  * divide n by one of its proper divisors, or 
  * subtract 1 from n if n is greater than 1. 



A proper divisor is a divisor of a number, excluding itself. For example, 1, 2, 4, 5, and 10 are proper divisors of 20, but 20 itself is not.

What is the minimum number of moves Ridbit is required to make to reduce n to 1?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The only line of each test case contains a single integer n (1 ≤ n ≤ 10^9).

Output

For each test case, output the minimum number of moves required to reduce n to 1.

Example

Input


6
1
2
3
4
6
9


Output


0
1
2
2
2
3

Note

For the test cases in the example, n may be reduced to 1 using the following operations in sequence

1

2 \xrightarrow{} 1

3 \xrightarrow{} 2 \xrightarrow{} 1

4 \xrightarrow{} 2 \xrightarrow{} 1

6 \xrightarrow{} 2 \xrightarrow{} 1

9 \xrightarrow{} 3 \xrightarrow{} 2\xrightarrow{} 1
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
 
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
 
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
 
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
 
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
 
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline()
 
# ------------------------------


def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def print_list(l):
    print(' '.join(map(str,l)))
# sys.setrecursionlimit(100000)
# from heapq import *
# from collections import deque as dq
# from math import ceil,floor,sqrt,pow,gcd,log
# import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc 
# from functools import reduce
# from functools import lru_cache

for _ in range(N()):
    n = N()
    if n==1:
        print(0)
    elif n==2:
        print(1)
    elif n==3:
        print(2)
    elif n&1==0:
        print(2)
    else:
        print(3)
