One day Vasya got hold of information on the Martian dollar course in bourles for the next n days. The buying prices and the selling prices for one dollar on day i are the same and are equal to ai. Vasya has b bourles. He can buy a certain number of dollars and then sell it no more than once in n days. According to Martian laws, one can buy only an integer number of dollars. Which maximal sum of money in bourles can Vasya get by the end of day n?

Input

The first line contains two integers n and b (1 ≤ n, b ≤ 2000) — the number of days and the initial number of money in bourles. The next line contains n integers ai (1 ≤ ai ≤ 2000) — the prices of Martian dollars.

Output

Print the single number — which maximal sum of money in bourles can Vasya get by the end of day n.

Examples

Input

2 4
3 7


Output

8


Input

4 10
4 3 2 1


Output

10


Input

4 10
4 2 3 1


Output

15
#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 50;
const long long mod = 1e9 + 7;
const long long INF = 1e18;
long long dx[] = {-1, 0, 1, 0};
long long dy[] = {0, 1, 0, -1};
bool compare(const pair<long long, long long> &i,
             const pair<long long, long long> &j) {
  return i.second > j.second;
}
long long fact[N];
vector<bool> is_prime(100000, true);
vector<long long> prime;
void pre() {
  fact[0] = 1;
  for (long long i = 1; i < N; i++) {
    fact[i] = i * fact[i - 1];
    fact[i] %= mod;
  }
}
long long modexp(long long x, long long n) {
  long long r = 1;
  while (n) {
    if (n & 1) r = (r * x) % mod;
    x = (x * x) % mod;
    n >>= 1;
  }
  return r % mod;
}
long long modinv(long long x) { return modexp(x, mod - 2); }
long long ncr(long long n, long long r) {
  if (n < r) return 0;
  long long ans =
      (fact[n] % mod * modinv(fact[r]) % mod * modinv(fact[n - r]) % mod) % mod;
  return ans;
}
long long exp(long long x, long long n) {
  long long r = 1;
  while (n) {
    if (n & 1) r = (r * x);
    x = (x * x);
    n >>= 1;
  }
  return r;
}
long long parent[N];
long long sz[N];
long long find_set(long long v) {
  if (v == parent[v]) return v;
  return parent[v] = find_set(parent[v]);
}
void make_set(long long v) {
  parent[v] = v;
  sz[v] = 1;
}
void union_sets(long long a, long long b) {
  a = find_set(a);
  b = find_set(b);
  if (a != b) {
    if (sz[a] < sz[b]) swap(a, b);
    parent[b] = a;
    sz[a] += sz[b];
  }
}
void solve() {
  long long n, b;
  cin >> n >> b;
  vector<long long> A(n);
  for (auto &x : A) cin >> x;
  long long mx = 0;
  for (long long i = 0; i < n; i++) {
    for (long long j = i; j < n; j++) {
      long long x = b / A[i];
      long long rem = b % A[i];
      long long ans = x * A[j] + rem;
      mx = max(mx, ans);
    }
  }
  cout << mx << endl;
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long i = 0; i < t; i++) {
    solve();
  }
  return 0;
}
