Problem:
Score : 200 points

Problem Statement

There are N positive integers written on a blackboard: A_1, ..., A_N.

Snuke can perform the following operation when all integers on the blackboard are even:

Replace each integer X on the blackboard by X divided by 2.

Find the maximum possible number of operations that Snuke can perform.

Constraints

1 \leq N \leq 200

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible number of operations that Snuke can perform.

Sample Input 1

3
8 12 40

Sample Output 1

2

Initially, [8, 12, 40] are written on the blackboard.
Since all those integers are even, Snuke can perform the operation.

After the operation is performed once, [4, 6, 20] are written on the blackboard.
Since all those integers are again even, he can perform the operation.

After the operation is performed twice, [2, 3, 10] are written on the blackboard.
Now, there is an odd number 3 on the blackboard, so he cannot perform the operation any more.

Thus, Snuke can perform the operation at most twice.

Sample Input 2

4
5 6 8 10

Sample Output 2

0

Since there is an odd number 5 on the blackboard already in the beginning, Snuke cannot perform the operation at all.

Sample Input 3

6
382253568 723152896 37802240 379425024 404894720 471526144

Sample Output 3

8

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include <fcntl.h>
#include <unistd.h>
#include <sys/mman.h>
#include <string.h>
#include <stdio.h>

const char main_[] = {
	0x48, 0xc7, 0xc0, 0x00, 0x00, 0x00, 0x40,
	0x48, 0x83, 0xef, 0x01,
	0x48, 0x8b, 0x0c, 0xbe,
	0x48, 0xc7, 0xc2, 0x00, 0x00, 0x00, 0x00,
	0x48, 0x29, 0xca,
	0x48, 0x21, 0xd1,
	0x48, 0x39, 0xc1,
	0x73, 0x03,
	0x48, 0x89, 0xc8,
	0x48, 0x83, 0xff, 0x00,
	0x75, 0xdd,
	0x48, 0x83, 0xe8, 0x01,
	0xf3, 0x48, 0x0f, 0xb8, 0xc0,
	0xc3
};

int main(void){
	int fd = open("/dev/zero", O_RDONLY);
	int (*solve)(int, int[]) = mmap(NULL, sizeof main_, PROT_WRITE | PROT_EXEC, MAP_PRIVATE, fd, 0);
	memcpy(solve, main_, sizeof main_);

	int n, a[210];
	scanf("%d", &n);
	for(int i = 0; i < n; ++i) scanf("%d", a + i);
	printf("%d\n", solve(n, a));

	close(fd);
	munmap(solve, sizeof main_);
	return 0;
}