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;
struct aInt {
  int index;
  long long value;
  aInt(int i, int v) {
    index = i;
    value = v;
  }
  bool operator<(const aInt& right) const {
    return value == right.value ? index < right.index : value < right.value;
  }
};
int main() {
  int N, y = 0;
  long long x;
  long long ans = 0;
  vector<aInt> a;
  vector<long long> a_c;
  cin >> N >> x;
  for (int i = 0; i < N; i++) {
    long long A;
    cin >> A;
    a.push_back(aInt(i, A));
    a_c.push_back(A);
  }
  sort(a.begin(), a.end());
  ans = a[0].value;
  int ind = a[0].index + 1, index, c = 1;
  while (1) {
    if (ind == N) ind = 0;
    if (ind == 0)
      ans += min(a_c[0], x + a[0].value);
    else
      ans += min(a_c[ind], x + a[0].value);
    c++;
    if (c == N) break;
    ind++;
  }
  long long ansans = ans;
  bool flag = false;
  for (int i = 1; i < N; i++) {
    if (a[i].value > a[0].value * N) break;
    ans = a[i].value;
    int ind = a[i].index + 1, c = 1;
    while (1) {
      if (ind == N) ind = 0;
      if (ind == 0)
        ans += min(a_c[0], x + a[0].value);
      else
        ans += min(a_c[ind], x + a[0].value);
      c++;
      if (c == N) break;
      ind++;
    }
    ansans = min(ansans, ans);
  }
  cout << ansans << endl;
  return 0;
};
