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 <bits/stdc++.h>
using namespace std;
 
//long long
using ll = long long;
 
// pair<int, int>
using PII = pair<int, int>;
 
//最大値、mod
const int MOD = 1000000007;
const int INF = 1000000000;
const long long LINF = 1e18;
const int MAX = 510000;
 
//出力系
#define print(x) cout << x << endl
#define prints(x) cout << fixed << setprecision(10) << x << endl
#define printc(x) cout << setw(2) << setfill('0') << x << endl;
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
 
// begin() end()
#define all(x) (x).begin(),(x).end()
 
//for
#define REP(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
 
//最大公約数 
unsigned gcd(unsigned a, unsigned b) {
  if(a < b) return gcd(b, a);
  unsigned r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}
 
// 最小公倍数
unsigned lcm(unsigned a, unsigned b){
    return a / gcd(a, b) * b; 
}
 
// a = max(a, b), a = min(a, b)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
 
// 階乗(MODをとる)
ll pow_mod(ll num, ll pow, ll mod) {
    ll prod = 1;
    num %= mod;
    while (pow > 0) {
        if (pow & 1) prod = prod * num % mod;
        num = num * num % mod;
        pow >>= 1;
    }
    return prod;
}
 
 
 
// 二項係数(MODとる、1 ≦ k ≦ n ≦ 10^7 程度)
// COMinit()
// COM(x, y)
// とコンビで使う
 
// テーブルを作る前処理
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 
//int dx[4] = {0, 0, 1, -1}, dy[4] = {1, -1, 0, 0};
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};

int main(){
    ll N, M;
    cin >> N >> M;
    vector<ll>a;
    for(int i = 1; i * i <= M; i++) if(M % i == 0) a.push_back(i);

    ll ans = 1;

    REP(i, a.size()) if(N * a.at(i) <= M) chmax(ans, a.at(i));
    
        

    print(ans);
    return 0;
    
}