#include <bits/stdc++.h>
#include <atcoder/all>
#include <iostream>
#include <cmath>
using namespace std;
using namespace atcoder;
using ll = long long;
using lb = long double;
using P = pair<ll, ll>;
using T = tuple<ll, ll, ll>;
using vll = vector<ll>;
using vb = vector<bool>;
using vvll = vector<vector<ll>>;
using vP = vector<P>;
using Graph = vector<vector<ll>>;
using WGraph = vector<vector<pair<ll, ll>>>; // コスト、頂点番号の順
using mint = modint998244353;
//using mint = long double;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
mt19937_64 rng(58);
long double PI = 3.14159265358979;
const ll LLMAX = 9223372036854775807;
const ll INF = 1e18;
vector<ll> di = {-1, 0, 1, 0}; // 上左下右
vector<ll> dj = {0, -1, 0, 1};
template<typename T> inline bool chmax(T &a, T b) { return ((a < b) ? (a = b, true) : (false)); }
template<typename T> inline bool chmin(T &a, T b) { return ((a > b) ? (a = b, true) : (false)); }

ll GCD(ll a, ll b) {
    while (a >= 1 && b >= 1) {
        if (a >= b) a = a % b;
        else b = b % a;
    }
    if (a != 0) return a;
    return b;
}

int main() {
    ll n, k;
    cin >> n >> k;
    vll a(n), b(n);
    rep (i, n) cin >> a[i];
    rep (i, n) cin >> b[i];

    vvll divA(n);
    rep (i, n) {
        for (ll div = 1; div * div <= a[i]; div++) {
            if (a[i] % div == 0) {
                divA[i].push_back(div);
                divA[i].push_back(a[i] / div);
            }
        }
        sort(divA[i].rbegin(), divA[i].rend());
    }

    priority_queue<P, vector<P>, greater<P>> q;
    rep (i, n) {
        q.push({gcd(a[i], b[i]), i});
    }

    ll ans = 1;
    while (k >= 0) {
        auto [pg, pi] = q.top(); q.pop();
        ans = pg;

        ll div = 0;
        // backで参照する時注意
        while (!divA[pi].empty() && pg >= divA[pi].back()) divA[pi].pop_back();
        if (divA[pi].empty()) break;
        div = divA[pi].back();

        // 普通にこれでいい、次へ進めるのはこれで簡単にいける
        k -= div - b[pi] % div;
        b[pi] += div - b[pi] % div;
        if (divA[pi].size() != 0) div = gcd(a[pi] / div, b[pi] / div) * div;
        
        if (k < 0) break;
        if (divA[pi].size() != 0) q.push({div, pi});
    }

    cout << ans << endl;
    return 0;
    // 全てをb[i]が全て1、a[i]が10^5以下の数で一番約数の数が多い数、にすると落ちるかも？
}