Problem:
Score : 400 points

Problem Statement

You are given integers N and M.

Consider a sequence a of length N consisting of positive integers such that a_1 + a_2 + ... + a_N = M. Find the maximum possible value of the greatest common divisor of a_1, a_2, ..., a_N.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

N \leq M \leq 10^9

Input

Input is given from Standard Input in the following format:

N M

Output

Print the maximum possible value of the greatest common divisor of a sequence a_1, a_2, ..., a_N that satisfies the condition.

Sample Input 1

3 14

Sample Output 1

2

Consider the sequence (a_1, a_2, a_3) = (2, 4, 8). Their greatest common divisor is 2, and this is the maximum value.

Sample Input 2

10 123

Sample Output 2

3

Sample Input 3

100000 1000000000

Sample Output 3

10000

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <math.h>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <string>
#include <cstring>
#include <regex>
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using pii = pair<int,int>;
using pl4 = pair<ll,ll>;
using vi = vector<int>;
using vvi = vector<vi>;
using vs = vector<string>;
using vvs = vector<vs>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vpii = vector<pii>;
using vvpii = vector<vpii>;
using vpl4 = vector<pl4>;
using vvpl4 = vector<vpl4>;

#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pob pop_back()
#define sz size()
#define be begin()
#define en end()
#define asn assign
#define emp empty()
#define ft front()
#define bk back()
#define clr clear()
#define ins insert
#define ers erase
#define res resize

#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define rFOR(i,a,b) for(int i=(b);i>=(a);i--)
#define REP(i,a) for(int i=0;i<(a);i++)
#define REP1(i,a) for(int i=1;i<=(a);i++)
#define rREP(i,a) for(int i=(a-1);i>=0;i--)
#define rREP1(i,a) for(int i=(a);i>0;i--)
#define SORT(a) sort((a).be,(a).en)
#define rSORT(a) sort((a).rbegin(),(a).rend())
#define UNIQUE(a) (a).erase(unique((a).be,(a).en),(a).en)
#define PREVP(a) prev_permutation((a).be,(a).en)
#define NEXTP(a) next_permutation((a).be,(a).en)
#define BINS(a,b) binary_search((a).be,(a).en,(b))
#define LOWB(a,b) (lower_bound((a).be,(a).en,(b))-(a).be)
#define UPB(a,b) (upper_bound((a).be,(a).en,(b))-(a).be)
#define CNT(a,b) count((a).be,(a).en,b)
#define SUM(a) accumulate((a).be,(a).en,0)
#define REV(a) reverse((a).be,(a).en)
#define REGS(a,b) regex_search((a),regex(b))
#define REGM(a,b) regex_match((a),regex(b))
#define yn(a) cout <<((a)?"yes":"no")<<endl;
#define Yn(a) cout <<((a)?"Yes":"No")<<endl;
#define YN(a) cout <<((a)?"YES":"NO")<<endl;
#define say(a) cout <<(a);
#define sal(a) cout <<(a)<<endl;
#define sak cout <<endl;
#define dbg(a) cout <<(#a)<<": "<<(a)<<endl;
#define a2l(a) ((ll)(a-97))
#define A2l(a) ((ll)(a-65))
#define l2a(a) ((char)(a+97))
#define l2A(a) ((char)(a+65))
#define DigN2(a) ((llabs(a)==0)?(1):((ll)(log2(double(llabs(a))))+1))
#define DigN10(a) ((llabs(a)==0)?(1):((ll)(log10(double(llabs(a))))+1))
#define Dig2(a,b) (((a)>>(b))&1)
#define Dig10(a,b) (ll)(((a)/((ll)(pow(10.0,(double)(b)))))%10)
#define Pow2(a) (1<<(a))
#define llin(a) ll (a);cin >>(a);
#define stin(a) string (a);cin >>(a);
#define rdn(a,b) ((a)/(b))
#define rou(a,b) ((((double(a)/double(b))-((a)/(b)))<0.5)?((a)/(b)):(((a)/(b))+1))
#define rup(a,b) ((((a)%(b))==0)?((a)/(b)):(((a)/(b))+1))
#define min(a,b) ((a<b)?(a):(b))
#define max(a,b) ((a>b)?(a):(b))

#define int ll

const ll MOD = 1e9+7;
const string alp = "abcdefghijklmnopqrstuvwxyz";
const string ALP = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

ll gcd(ll a, ll b){if(b==0)return a;return gcd(b,a%b);}
ll lcm(ll a, ll b){return a/gcd(a,b)*b;}

ll sgn(ll n){
    if(n==0) return 0;
    else return n/abs(n);
}

ll salv(vll v){
  say("{");
  FOR(i,0,v.sz-1){
    say(v[i]);
    if(i!=v.sz-1) say(",");
  }
  sal("}")
}

ll DigS10(ll n){
  ll m=0;
  FOR(i,0,DigN10(n)-1){
    m+=(ll)((llabs(n)%(ll)(pow(10.0,(double)(i+1))))/(ll)(pow(10.0,(double)i)));
  }
  return m;
}

ll isP(ll n){
  if(n<=1) return 0;
  FOR(i,2,(ll)sqrt(n)){
    if(n%i==0) return 0;
  }
  return 1;
}

vll FactM(1,1);
vll FactMI(1,1);

ll PowM(ll a,ll b){
  ll ans=1,x=(a%MOD);
  FOR(i,0,DigN2(b)-1){
    if(Dig2(b,i)==1) ans=(ans*x)%MOD;
    if(i!=(DigN2(b)-1)) x=(x*x)%MOD;
  }
  return ans;
}

void CFactM(ll n){
  FOR(i,FactM.sz,n){
    FactM.pb((FactM[i-1]*(i%MOD))%MOD);
  }
  return;
}

void CFactMI(ll n){
  CFactM(n);
  if(FactMI.sz<(n+1)) FactMI.res(n+1,-1);
  if(FactMI[n]==-1) FactMI[n]=PowM(FactM[n],MOD-2);
  rFOR(i,1,n-1){
    if(FactMI[i]!=-1) break;
    FactMI[i]=((FactMI[i+1]*((i+1)%MOD))%MOD);
  }
  return;
}

ll CombM(ll n,ll k){
    if((n<0)||(k<0)) return 0;
    if(n<k) return 0;
    if(n+1>FactMI.sz) CFactMI(n);
    return ((((FactMI[k]*FactMI[n-k])%MOD)*FactM[n])%MOD);
}

signed main() {
  ll n,m,r=1;
  cin >> n >> m;
  if (n==1){
    cout << m;
    return 0;
  }
  if(n*n>m){
    FOR(i,1,n){
    if(m%i==0){
      r=i;
      
    }
    if(m/i<n)i=n+1;
  }
  cout << r;
      return 0;
  } else if (n*n<m){
    FOR(i,1,m/n){
    if(m%i==0){
      r=i;
    }
  }
  cout << r;
      return 0;
  } else {
    cout << n;
    return 0;
  }
  cout<<1;
  return 0;
}