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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e15);
const ll mod = 1000000007;
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
  __INIT() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
} __init;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
int main() {
  ll n, x;
  cin >> n >> x;
  V<ll> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  V<P> res(n, {inf, inf});
  ll ans = inf;
  for (ll i = 0; i < n; i++) {
    for (ll j = 0; j < n; j++) {
      int id = (i + j) % n;
      if (res[id].first + res[id].second > a[i] + j * x) {
        res[id].second = j * x;
        res[id].first = a[i];
      }
      if (res[id].first + res[id].second == a[i] + j * x &&
          res[id].second > j * x) {
        res[id].second = j * x;
        res[id].first = a[i];
      }
    }
  }
  ll sum = 0, tmp = 0;
  for (ll j = 0; j < n; j++) {
    chmax(tmp, res[j].second);
    sum += res[j].first;
  }
  cout << sum + tmp << "\n";
}
