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;
template <class T>
inline void remax(T& A, T B) {
  if (A < B) A = B;
}
template <class T>
inline void remin(T& A, T B) {
  if (A > B) A = B;
}
string ToString(long long num) {
  string ret;
  do {
    ret += ((num % 10) + '0');
    num /= 10;
  } while (num);
  reverse(ret.begin(), ret.end());
  return ret;
}
long long ToNumber(string s) {
  long long r = 0, p = 1;
  for (int i = s.size() - 1; i >= 0; --i) r += (s[i] - '0') * p, p *= 10;
  return r;
}
long long Gcd(long long a, long long b) {
  while (a %= b ^= a ^= b ^= a)
    ;
  return b;
}
long long Power(long long base, long long power) {
  long long ret = 1;
  while (power) {
    if (power & 1) ret *= base;
    power >>= 1;
    base *= base;
  }
  return ret;
}
long long PowerMod(long long base, long long power, long long mod) {
  if (!power) return 1;
  if (power & 1) return (base * PowerMod(base, power - 1, mod)) % mod;
  return PowerMod((base * base) % mod, power >> 1, mod);
}
int Log(long long num, long long base) {
  int ret = 0;
  while (num) {
    ++ret;
    num /= base;
  }
  return ret;
}
int Count(long long mask) {
  int ret = 0;
  while (mask) {
    if (mask & 1) ++ret;
    mask >>= 1;
  }
  return ret;
}
inline void run() {
  int arr[2000];
  int n, b, ans;
  scanf("%d%d", &n, &b);
  for (int i = 0; i < n;) scanf("%d", arr + i++);
  ans = b;
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j)
      remax(ans, arr[j] * (b / arr[i]) + b % arr[i]);
  printf("%d\n", ans);
}
int main() {
  run();
  return 0;
}
