// Original C++ solution (slower):
#include <bits/stdc++.h>

#define FOR(i,a,b) for (int i=(a);i<(b);i++)

#define FORR(i,a,b) for (int i=(a);i>=(b);i--)

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define pcnt __builtin_popcount

#define sz(x) (int)(x).size()

#define maxs(x,y) x = max(x,y)

#define mins(x,y) x = min(x,y)

#define show(x) cout<<#x<<" = "<<x<<endl;

#define rng(a) (a.begin()),(a.end())

#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)



using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

typedef vector<int> vi;

typedef vector<vi> vvi;

typedef vector<pii> vpii;

typedef set<int> si;

typedef pair<ll,ll> pll;

typedef vector<ll> vl;

typedef vector<vl> vvl;

typedef vector<pll> vpll;

typedef set<ll> sl;

typedef __int128_t lll;

typedef pair<lll,lll> plll;

typedef vector<lll> vlll;

template<typename T>string join(vector<T>&v){stringstream s;FOR(i,0,sz(v))s<<' '<<v[i];return s.str().substr(1);}

template<typename A,size_t N,typename T>void Fill(A (&array)[N],const T&v){fill((T*)array,(T*)(array+N),v);}

lll gcd(lll a,lll b,lll &x,lll &y){if(!b){x=1;y=0;return a;}lll d=gcd(b,a%b,y,x);y-=a/b*x;return d;}

ll gcd(ll a,ll b){lll x=0,y=0;return gcd(a,b,x,y);}

ll modpow(lll a,lll n,ll m){if(a==0)return a;lll p=1;for(;n>0;n<<=1,a=a*a%m)if(n&1)p=p*a%m;return(ll)p;}

void dout(double d){printf("%.12f\n",d);}

void YN(bool z){cout<<(z?"YES\n":"NO\n");}

void Yn(bool z){cout<<(z?"Yes\n":"No\n");}

void yn(bool z){cout<<(z?"yes\n":"no\n");}



const int iinf = 1e9;

const ll linf = 1e18;

const int mod = 1e9+7;

const double pi = acos(-1);

const double eps = 1e-10;

const int N = 0;

struct Prime{

  vector<ll> pl;

  void init(int a){ // a以下の素数列挙

    pl.resize(0);

    vector<bool> z(a+1, true);

    FOR(i, 2, a+1){

      if(z[i])pl.push_back(i);

      for(int b=i*2; b <= a; b+=i) z[b]=false;

    }

  }

  void divisor(ll a, vector<ll>&v){ // v に約数が入る

    v.resize(0);

    v.push_back(1);

    for(ll b: pl){

      if(b*b > a) break;

      if(a%b) continue;

      int l=0, vs=v.size();

      while(a%b==0){

        a /= b;

        FORR(i, v.size()-1, l) v.push_back(v[i] * b);

        l += vs;

      }

    }

    if(a>1) FORR(i, v.size()-1, 0) v.push_back(v[i] * a);

    sort(v.begin(), v.end());

  }

} pr;

ll n, m;

vl v;



main(){

  cin.tie(0);

  ios::sync_with_stdio(false);

  cin >> n >> m;

  pr.init(100000);

  pr.divisor(m, v);

  for(ll a: v)if(a >= n){

    cout << m/a << endl;

    return 0;

  }

}

// Optimized solution to the same problem (faster):
#include <bits/stdc++.h>

#define FOR(i,a,b) for (int i=(a);i<(b);i++)

#define FORR(i,a,b) for (int i=(a);i>=(b);i--)

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define pcnt __builtin_popcount

#define sz(x) (int)(x).size()

#define maxs(x,y) x = max(x,y)

#define mins(x,y) x = min(x,y)

#define show(x) cout<<#x<<" = "<<x<<endl;

#define rng(a) (a.begin()),(a.end())

#define each(it,c) for(__typeof((c).begin()) it=(c).begin();it!=(c).end();it++)



using namespace std;

typedef long long ll;

typedef pair<int,int> pii;

typedef vector<int> vi;

typedef vector<vi> vvi;

typedef vector<pii> vpii;

typedef set<int> si;

typedef pair<ll,ll> pll;

typedef vector<ll> vl;

typedef vector<vl> vvl;

typedef vector<pll> vpll;

typedef set<ll> sl;

typedef __int128_t lll;

typedef pair<lll,lll> plll;

typedef vector<lll> vlll;

template<typename T>string join(vector<T>&v){stringstream s;FOR(i,0,sz(v))s<<' '<<v[i];return s.str().substr(1);}

template<typename A,size_t N,typename T>void Fill(A (&array)[N],const T&v){fill((T*)array,(T*)(array+N),v);}

lll gcd(lll a,lll b,lll &x,lll &y){if(!b){x=1;y=0;return a;}lll d=gcd(b,a%b,y,x);y-=a/b*x;return d;}

ll gcd(ll a,ll b){lll x=0,y=0;return gcd(a,b,x,y);}

ll modpow(lll a,lll n,ll m){if(a==0)return a;lll p=1;for(;n>0;n<<=1,a=a*a%m)if(n&1)p=p*a%m;return(ll)p;}

void dout(double d){printf("%.12f\n",d);}

void YN(bool z){cout<<(z?"YES\n":"NO\n");}

void Yn(bool z){cout<<(z?"Yes\n":"No\n");}

void yn(bool z){cout<<(z?"yes\n":"no\n");}



const int iinf = 1e9;

const ll linf = 1e18;

const int mod = 1e9+7;

const double pi = acos(-1);

const double eps = 1e-10;

const int N = 0;

struct Prime{

  vector<ll> pl;

  void init(int a){ // a以下の素数列挙

    pl.resize(0);

    vector<bool> z(a+1, true);

    FOR(i, 2, a+1){

      if(z[i])pl.push_back(i);

      for(int b=i*2; b <= a; b+=i) z[b]=false;

    }

  }

  void divisor(ll a, vector<ll>&v){ // v に約数が入る

    v.resize(0);

    v.push_back(1);

    for(ll b: pl){

      if(b*b > a) break;

      if(a%b) continue;

      int l=0, vs=v.size();

      while(a%b==0){

        a /= b;

        FORR(i, v.size()-1, l) v.push_back(v[i] * b);

        l += vs;

      }

    }

    if(a>1) FORR(i, v.size()-1, 0) v.push_back(v[i] * a);

    sort(v.begin(), v.end());

  }

} pr;

ll n, m;

vl v;



main(){

  cin.tie(0);

  ios::sync_with_stdio(false);

  cin >> n >> m;

  pr.init(32000);

  pr.divisor(m, v);

  for(ll a: v)if(a >= n){

    cout << m/a << endl;

    return 0;

  }

}

// Fastest known solution for this problem: