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;
template <class T>
bool chmax(T &a, const T &b) {
  return (a < b) ? (a = b, 1) : 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  return (b < a) ? (a = b, 1) : 0;
}
using ll = long long;
using R = long double;
const R EPS = 1e-9;
inline int sgn(const R &r) { return (r > EPS) - (r < -EPS); }
inline R sq(R x) { return sqrt(max<R>(x, 0.0)); }
const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const ll inf = 1LL << 60;
ll a[4010];
int main(void) {
  ll n, x;
  cin >> n >> x;
  for (int i = int(0); i < int(n); ++i) cin >> a[i];
  for (int i = int(0); i < int(n); ++i) a[i + n] = a[i];
  ll ans = 0LL, cmax = 0LL;
  for (int i = int(0); i < int(n); ++i) {
    ll cmin = inf, add = -1, num = -1;
    for (int j = int(0); j < int(n); ++j)
      if (chmin(cmin, a[n + i - j] + x * j)) add = a[n + i - j], num = j;
    chmax(cmax, num);
    ans += add;
  }
  cout << ans + cmax * x << endl;
  return 0;
}
