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>
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;
}
const int vx[4] = {0, 1, 0, -1}, vy[4] = {1, 0, -1, 0};
using namespace std;
int main() {
  long long N, X, ans = 0LL, t, e = ((1LL << 62) - (1LL << 31));
  cin >> N >> X;
  long long A[3 * N + 1];
  for (int i = 1; i <= N; ++i) {
    cin >> A[i];
    A[N + i] = A[i];
    A[2 * N + i] = A[i];
    if (e > A[i]) {
      e = A[i];
      t = i;
    }
  }
  vector<pair<long long, long long>> L;
  long long tt = t, v = t;
  while (t < tt + N) {
    while (A[v] + X * (t + 1 - v) <= A[t + 1] && t < tt + N - 1) {
      t++;
    }
    L.emplace_back(make_pair(A[v], t - v + 1));
    v = t + 1;
    t = v;
  }
  long long mx = 0LL;
  for (int i = (0); i < (L.size()); ++i) {
    chmax(mx, L[i].second);
  }
  for (int i = (0); i < (L.size()); ++i) {
    ans += L[i].first * L[i].second;
  }
  cout << ans + X * (mx - 1) << '\n';
  return 0;
}
