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 long long INF = 1e12;
const long long MOD = 1e9 + 7;
long long sq(long long num) { return num * num; }
long long mod_pow(long long x, long long n) {
  if (n == 0) return 1;
  if (n == 1) return x % MOD;
  long long res = sq(mod_pow(x, n / 2));
  res %= MOD;
  if (n % 2 == 1) {
    res *= x;
    res %= MOD;
  }
  return res;
}
long long mod_add(long long a, long long b) { return (a + b) % MOD; }
long long mod_sub(long long a, long long b) { return (a - b + MOD) % MOD; }
long long mod_mul(long long a, long long b) { return a * b % MOD; }
int n, x;
int a[100100];
int main() {
  cin >> n >> x;
  for (int i = 0; i < (n); i++) cin >> a[i];
  long long ans = INF;
  for (int i = 0; i < (n); i++) {
    long long sum = 0;
    long long now = i;
    long long cnt = 1;
    long long w = 1;
    for (int j = 0; j < (n); j++) {
      sum += a[now];
      if (a[now] + cnt * x <= a[(now + cnt) % n])
        cnt++;
      else {
        now = (now + cnt) % n;
        cnt = 1;
      }
      w = max(w, cnt);
    }
    ans = min(sum + (w - 1) * x, ans);
  }
  cout << ans << endl;
}
