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;
long long gcd(long long aaa, long long bbb) {
  if (bbb == 0) {
    return aaa;
  }
  return gcd(bbb, aaa % bbb);
}
int main() {
  long long int N, x;
  cin >> N >> x;
  long long int a[N], s = 0, cmp, tmp[2000] = {}, l = 0, count = -1, o,
                      countt = 0, n[2000] = {}, tmpp[2000] = {};
  map<long long int, long long int> m;
  map<long long int, long long int> c;
  for (int i = 0; i < N; i++) {
    tmpp[i] = 1e9;
  }
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    s += a[i];
    if (a[i] < x) {
      m[i]++;
    }
  }
  for (int i = 0; i < N * 2; i++) {
    if (i == N && count == 0) {
      break;
    } else if (i == N) {
      countt++;
    }
    o = i % N;
    if (count == -1) {
      if (m[i] == 1) {
        l = i;
        count++;
      }
    } else {
      if (m[i] == 1) {
        n[count] = abs(o - l) * a[l];
        count++;
        l = i;
        if (countt == 1) {
          break;
        }
      }
    }
  }
  sort(n, n + 2000);
  long long int ans = s;
  for (int i = 0; i < N; i++) {
    tmp[i] = a[i];
    for (int j = 0; j < N; j++) {
      cmp = i + 1;
      cmp += j;
      if (cmp >= N) {
        cmp -= N;
      }
      if (i == cmp) {
        break;
      }
      tmp[i] += min(a[cmp], a[i] + x);
    }
  }
  countt = count + 1;
  if (count >= 2) {
    for (int i = 0; i < N; i++) {
      if (m[i] == 1) {
        for (int i = 1; i <= count; i++) {
          tmpp[i] += n[count - i];
        }
      } else {
        break;
      }
    }
  }
  for (int i = 0; i < N; i++) {
    tmp[i] = min(tmp[i], tmpp[i]);
  }
  for (int i = 0; i < N; i++) {
    ans = min(tmp[i], ans);
  }
  cout << ans << endl;
  return 0;
}
