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;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vs = vector<string>;
using pll = pair<ll, ll>;
using vp = vector<pll>;
const ll MOD = 1000000007;
const ll INF = 100000000000000000LL;
inline ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
inline ll LCM(ll a, ll b) { return a / GCD(a, b) * b; }
inline ll powint(ll x, ll y) {
  ll r = 1;
  while (y) {
    if (y & 1) r *= x;
    x *= x;
    y >>= 1;
  }
  return r;
}
inline ll powmod(ll x, ll y, ll m = MOD) {
  ll r = 1;
  while (y) {
    if (y & 1) r *= x;
    x *= x;
    r %= m;
    x %= m;
    y >>= 1;
  }
  return r;
}
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 (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(15);
  ll n, x;
  cin >> n >> x;
  vll a(n);
  for (ll i = 0; i < (n); i++) {
    cin >> a[i];
  }
  ll xop = 0;
  vll cost(n, INF);
  vll cost2(n, INF);
  for (ll i = 0; i < (n); i++) {
    ll index = 0;
    for (ll j = 0; j < (n); j++) {
      if (chmin(cost[i], a[j] + ((i - j + n) % n) * x)) {
        index = j;
        cost2[i] = a[j];
      }
    }
    chmax(xop, (i - index + n) % n);
  }
  cout << accumulate((cost2).begin(), (cost2).end(), 0ll) + xop * x << endl;
  return 0;
}
