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;
namespace {
template <typename T>
using min_pq = priority_queue<T, vector<T>, greater<T>>;
long long int expmod(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
struct IOS {
  IOS() {
    ios::sync_with_stdio(0);
    cin.tie(0);
  }
} IOS;
}  // namespace
const int MAX_N = 200000;
const long long int MODD = 1000000009LL;
const long double EPS = 1e-9;
int main() {
  int N, x;
  cin >> N >> x;
  vector<long long int> a(N);
  for (__typeof(N) i = (0) - ((0) > (N)); i != (N) - ((0) > (N));
       i += 1 - 2 * ((0) > (N)))
    cin >> a[i];
  vector<vector<long long int>> cost(N, vector<long long int>(N, LLONG_MAX));
  for (__typeof(N) i = (0) - ((0) > (N)); i != (N) - ((0) > (N));
       i += 1 - 2 * ((0) > (N)))
    for (__typeof(N) j = (0) - ((0) > (N)); j != (N) - ((0) > (N));
         j += 1 - 2 * ((0) > (N))) {
      if (j == 0)
        cost[i][j] = a[i];
      else {
        cost[i][j] = min(cost[i][j - 1], a[(i - j + N) % N]);
      }
    }
  long long int best = LLONG_MAX;
  for (__typeof(N) j = (0) - ((0) > (N)); j != (N) - ((0) > (N));
       j += 1 - 2 * ((0) > (N))) {
    long long int cur = j * x;
    for (__typeof(N) i = (0) - ((0) > (N)); i != (N) - ((0) > (N));
         i += 1 - 2 * ((0) > (N)))
      cur += cost[i][j];
    best = min(best, cur);
  }
  cout << best << endl;
}
