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 times, we get string b. For example, string "abab" has two divisors — "ab" and "abab".

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 s2.

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".
#include <bits/stdc++.h>
using namespace std;
void adskiy_razgon() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int MOD = 1e9 + 7;
const int MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = acos((long double)-1);
const int xd[4] = {1, 0, -1, 0}, yd[4] = {0, 1, 0, -1};
long long nod(long long a, long long b) {
  if (b > a) {
    swap(a, b);
  }
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long nok(long long a, long long b) { return a * b / nod(a, b); }
void sp(long long a, double b) { cout << fixed << setprecision(a) << b; }
long long binpow(long long a, long long n) {
  if (n == 0) {
    return 1;
  }
  if (n % 2 == 1) {
    return binpow(a, n - 1) * a;
  } else {
    long long b = binpow(a, n / 2);
    return b * b;
  }
}
template <class A>
void read(vector<A>& v);
template <class A, size_t S>
void read(array<A, S>& a);
template <class T>
void read(T& x) {
  cin >> x;
}
void read(double& d) {
  string t;
  read(t);
  d = stod(t);
}
void read(long double& d) {
  string t;
  read(t);
  d = stold(t);
}
template <class H, class... T>
void read(H& h, T&... t) {
  read(h);
  read(t...);
}
template <class A>
void read(vector<A>& x) {
  for (auto& a : x) read(a);
}
template <class A, size_t S>
void read(array<A, S>& x) {
  for (auto& a : x) read(a);
}
string to_string(char c) { return string(1, c); }
string to_string(bool b) { return b ? "true" : "false"; }
string to_string(const char* s) { return string(s); }
string to_string(string s) { return s; }
string to_string(vector<bool> v) {
  string res;
  for (int i = (0); (1) > 0 ? i < ((int)(v).size()) : i > ((int)(v).size());
       i += (1))
    res += char('0' + v[i]);
  return res;
}
template <size_t S>
string to_string(bitset<S> b) {
  string res;
  for (int i = (0); (1) > 0 ? i < (S) : i > (S); i += (1))
    res += char('0' + b[i]);
  return res;
}
template <class T>
string to_string(T v) {
  bool f = 1;
  string res;
  for (auto& x : v) {
    if (!f) res += ' ';
    f = 0;
    res += to_string(x);
  }
  return res;
}
template <class A>
void write(A x) {
  cout << to_string(x);
}
template <class H, class... T>
void write(const H& h, const T&... t) {
  write(h);
  write(t...);
}
void print() { write("\n"); }
template <class H, class... T>
void print(const H& h, const T&... t) {
  write(h);
  if (sizeof...(t)) write(' ');
  print(t...);
}
void yes(bool ok) { print((ok ? "YES" : "Yes")); }
void no(bool ok) { print((ok ? "NO" : "No")); }
bool check(string a, string b) {
  int x = (int)(b).size() / (int)(a).size();
  string w = a;
  x--;
  while (x) {
    a += w;
    x--;
  }
  return a == b;
}
void solve() {
  string a, b;
  read(a, b);
  string w = "";
  int ans = 0;
  for (int i = 0; i < (int)(a).size(); ++i) {
    w += a[i];
    if ((int)(b).size() % (int)(w).size() == 0) {
      int x = (int)(b).size() / (int)(w).size() - 1;
      string k = w;
      while (x) k += w, x--;
      if (k == b) {
        string k = w;
        while ((int)(k).size() <= (int)(a).size()) {
          if ((int)(a).size() % (int)(k).size() == 0 &&
              (int)(b).size() % (int)(k).size() == 0 && check(k, a))
            ans++;
          k += w;
        }
        break;
      }
    }
  }
  print(ans);
}
int main() {
  adskiy_razgon();
  long long t = 1;
  for (int i = (0); (1) > 0 ? i < (t) : i > (t); i += (1)) {
    solve();
  }
  return 0;
}
