#include <bits/stdc++.h>
using namespace std;

using ll = long long;

const int MAXN = 200005;
const int MAXK = 25;

int N;
ll X;
ll A[MAXN];

// divisors[X + k]
vector<ll> divisors[MAXK];

// cost[i][k]: cost to make A[i] gcd > 1 with X + k
ll cost[MAXN][MAXK];

// prefix sum for cost[][k]
ll pref[MAXK][MAXN];

// trial division up to sqrt(X)
void factorize(ll Y, vector<ll> &D) {
    for (ll d = 2; d * d <= Y; ++d) {
        if (Y % d == 0) {
            D.push_back(d);
            while (Y % d == 0) Y /= d;
        }
    }
    if (Y > 1) D.push_back(Y);
}

ll get_cost(ll a, const vector<ll> &D) {
    ll best = LLONG_MAX;
    for (ll d : D) {
        ll add = (d - a % d) % d;
        best = min(best, add);
    }
    return best;
}

ll get_sum(int k, int l, int r) {
    if (l == 0) return pref[k][r];
    return pref[k][r] - pref[k][l - 1];
}

ll solve(int l, int r, int k) {
    if (k >= MAXK) return LLONG_MAX; // exceeds limit
    if (l == r) return cost[l][k];

    int m = (l + r) / 2;

    ll left = solve(l, m, k + 1);
    ll right_gcd = get_sum(k, m + 1, r);

    ll right = solve(m + 1, r, k + 1);
    ll left_gcd = get_sum(k, l, m);

    return min(left + right_gcd, right + left_gcd);
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> N >> X;
    for (int i = 0; i < N; ++i) {
        cin >> A[i];
    }

    for (int k = 0; k < MAXK; ++k) {
        ll Y = X + k;
        factorize(Y, divisors[k]);
    }

    for (int i = 0; i < N; ++i) {
        for (int k = 0; k < MAXK; ++k) {
            cost[i][k] = get_cost(A[i], divisors[k]);
        }
    }

    for (int k = 0; k < MAXK; ++k) {
        pref[k][0] = cost[0][k];
        for (int i = 1; i < N; ++i) {
            pref[k][i] = pref[k][i - 1] + cost[i][k];
        }
    }

    ll ans = solve(0, N - 1, 0);
    if (ans >= LLONG_MAX / 2) cout << -1 << '\n';
    else cout << ans << '\n';

    return 0;
}