Vasya has recently learned at school what a number's divisor is and decided to determine a string's divisor. Here is what he came up with.

String a is the divisor of string b if and only if there exists a positive integer x such that if we write out string a consecutively x

Now Vasya wants to write a program that calculates the number of common divisors of two strings. Please help him.

Input

The first input line contains a non-empty string s1.

The second input line contains a non-empty string s

Lengths of strings s1 and s2 are positive and do not exceed 105. The strings only consist of lowercase Latin letters.

Output

Print the number of common divisors of strings s1 and s2.

Examples

Input

abcdabcd
abcdabcdabcdabcd

Output

2

Input

aaa
aa

Output

1

Note

In first sample the common divisors are strings "abcd" and "abcdabcd".

In the second sample the common divisor is a single string "a". String "aa" isn't included in the answer as it isn't a divisor of string "aaa".

青島邀請賽的題，一直沒補，學完kmp再補上

先用kmo處理出兩個串的最小迴圈節，再分情況討論

一開始wa了一遍因為忽略了兩個串相同的情況

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxn 100005
using namespace std;
char t[maxn];
char p[maxn];
int f[maxn];
int n,m;
void getfail1()
{
    f[0]=f[1]=0;
    for(int i=1;i<n;i++)
    {
        int j=f[i];
        while(j&&t[i]!=t[j])
            j=f[j];
        f[i+1]=(t[i]==t[j])?j+1:0;
    }
}
void getfail2()
{
    f[0]=f[1]=0;
    for(int i=1;i<m;i++)
    {
        int j=f[i];
        while(j&&p[i]!=p[j])
            j=f[j];
        f[i+1]=(p[i]==p[j])?j+1:0;
    }
}
int main()
{scanf("%s%s",t,p);
n=strlen(t);
getfail1();
int len1=n-f[n];
if(len1==0||n%len1)
   len1=n;

m=strlen(p);
getfail2();
int len2=m-f[m];
if(len2==0||m%len2)
    len2=m;
if(len1!=len2||n%len1!=0||m%len2!=0)
    printf("0\n");
else
{
    int ans=0;
    for(int i=0;i<len1;i++)
        if(t[i]!=p[i])
    {printf("0\n");
return 0;
}
for(int i=len1;i<=n&&i<=m;i+=len1)
{
    if(n%i==0&&m%i==0)
        ans++;
}
    printf("%d\n",ans);
}
return 0;
}