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>
const int MOD = 1e9 + 7;
const int iINF = 2147483647 / 2;
const long long int llINF = 9223372036854775807 / 2;
using namespace std;
using ll = long long int;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
using vvvl = vector<vector<vector<ll>>>;
bool paircomp(const pair<ll, ll> &a, const pair<ll, ll> &b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first < b.first;
}
ll POW(ll n, ll m) {
  if (m == 0) {
    return 1;
  } else if (m % 2 == 0) {
    ll tmp = POW(n, m / 2);
    return (tmp * tmp);
  } else {
    return (n * POW(n, m - 1));
  }
}
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll N, x;
  cin >> N >> x;
  vl A(N);
  for (ll i = 0; i < (N); i++) cin >> A[i];
  ll ans = 0;
  ll maxloop = 0;
  for (ll i = 0; i < (N); i++) {
    ll score = llINF;
    ll loop = 0;
    for (ll j = 0; j < (N); j++) {
      ll po = i - j;
      if (j > i) {
        po = i + N - j;
      }
      if (score > A[j] + (po * x)) {
        score = A[j] + (po * x);
        loop = po * x;
      }
    }
    ans += score - loop;
    maxloop = max(maxloop, loop);
  }
  cout << ans + maxloop << endl;
  return 0;
}
