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;
const ll MOD = pow(10, 9) + 7;
ll mod(ll A, ll M) { return (A % M + M) % M; }
const ll INF = 1LL << 60;
template <class T>
bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
ll divCeil(ll A, ll B) { return (A + (B - 1)) / B; }
ll myctoi(char C) { return C - '0'; }
char myitoc(ll N) { return '0' + N; }
int main() {
  ll N, x;
  cin >> N >> x;
  vector<ll> A0(N);
  for (ll i = 0; i < N; i++) {
    cin >> A0.at(i);
  }
  ll minind = distance(A0.begin(), min_element(A0.begin(), A0.end()));
  vector<ll> A, B;
  for (ll i = minind; i < minind + N; i++) {
    A.push_back(A0.at(i % N));
    B.push_back(A0.at(i % N));
  }
  for (ll i = 1; i < N; i++) {
    if (A.at(i - 1) == B.at(i - 1))
      chmin(A.at(i), A.at(i - 1) + x);
    else
      chmin(A.at(i), A.at(i - 1));
  }
  ll ans = accumulate(A.begin(), A.end(), (ll)0);
  cout << ans << endl;
}
