Snuke lives in another world, where slimes are real creatures and kept by some people. Slimes come in N colors. Those colors are conveniently numbered 1 through N. Snuke currently has no slime. His objective is to have slimes of all the colors together.

Snuke can perform the following two actions:

* Select a color i (1≤i≤N), such that he does not currently have a slime in color i, and catch a slime in color i. This action takes him a_i seconds.

* Cast a spell, which changes the color of all the slimes that he currently has. The color of a slime in color i (1≤i≤N-1) will become color i+1, and the color of a slime in color N will become color 1. This action takes him x seconds.




Find the minimum time that Snuke needs to have slimes in all N colors.

Constraints

* 2≤N≤2,000
* a_i are integers.
* 1≤a_i≤10^9
* x is an integer.
* 1≤x≤10^9

Input

The input is given from Standard Input in the following format:


N x
a_1 a_2 ... a_N


Output

Find the minimum time that Snuke needs to have slimes in all N colors.

Examples

Input

2 10
1 100


Output

12


Input

3 10
100 1 100


Output

23


Input

4 10
1 2 3 4


Output

10
#include <bits/stdc++.h>
using namespace std;
const int maxn = int(1.5 * 1e7);
const int inf = 1e9 + 7;
const long long ll_inf = 1e18 + 420;
const double eps = 1e-4;
const int N = 1e6 + 7;
const int MAX = 2e5 + 9;
const int MOD = 998244353;
const long double pi = 3.14159265359;
long long fast_pow(int a, int b) {
  int mod = 1e9 + 7;
  long long res = a, ret = 1;
  while (b > 0) {
    if (b % 2) ret = (ret * res) % mod;
    res = (res * res) % mod;
    b /= 2;
  }
  return ret;
}
long long fact(long long n) {
  int mod = 1e9 + 7;
  if (n == 1 || n == 0) return 1;
  return ((n % mod) * (fact(n - 1) % mod) % mod);
}
long long Cnk(long long n, long long r) {
  int mod = 1e9 + 7;
  return (fact(n) * fast_pow((fact(r) * fact(n - r)) % mod, mod - 2)) % mod;
}
int n, x, a[5000];
long long dp[5000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; ++i) {
    dp[i] = dp[i - 1] + min(x + a[1], a[i]);
  }
  cout << dp[n];
}
