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<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#define max(x,y)(x>y?x:y)
#define min(x,y)(x>y?y:x)
typedef long long int ll;
typedef unsigned long long int ull;
int asc_cmp(const void* X,const void* Y){int x=*(int*)X,y=*(int*)Y;if(x<y)return -1;else if(x==y)return 0;else return 1;}
int desc_cmp(const void* X,const void* Y){int x=*(int*)X,y=*(int*)Y;if(x<y)return 1;else if(x==y)return 0;else return -1;}

int main(){
    int n;scanf("%d",&n);
    int a[n];for(int i=0;i<n;i++)scanf("%d",&a[i]);
    int ans=0;
    while(1){
        for(int i=0;i<n;i++){
            if(a[i]%2)goto END;
            a[i]/=2;
        }
        ans++;
    }
    END:

    printf("%d\n",ans);
}