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;
const bool debug = 0;
int N, x;
long long a[2016];
long long mini[2016];
bool fin[2016];
const int IINF = INT_MAX;
class RMQ {
 private:
  int limit, N;
  vector<int> dat;

 public:
  void init(int n_) {
    N = n_;
    limit = 1;
    while (limit < n_) limit *= 2;
    dat.clear();
    dat.resize(2 * limit, IINF);
  }
  int _build(int cur, int L, int R, const vector<int> &buf) {
    if (!(0 <= cur && cur < 2 * limit)) return IINF;
    if (L == R - 1) {
      if (L >= N) return IINF;
      dat[cur] = buf[L];
    } else {
      int vl = _build(cur * 2 + 1, L, (L + R) / 2, buf);
      int vr = _build(cur * 2 + 2, (L + R) / 2, R, buf);
      dat[cur] = min(vl, vr);
    }
    return dat[cur];
  }
  void build(const vector<int> &buf) { _build(0, 0, limit, buf); };
  void update(int k, int a) {
    k += limit - 1;
    dat[k] = a;
    while (k > 0) {
      k = (k - 1) / 2;
      dat[k] = min(dat[k * 2 + 1], dat[k * 2 + 2]);
    }
  }
  int _query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l)
      return IINF;
    else if (a <= l && r <= b)
      return dat[k];
    int vl = _query(a, b, k * 2 + 1, l, (l + r) / 2);
    int vr = _query(a, b, k * 2 + 2, (l + r) / 2, r);
    return min(vl, vr);
  }
  int query(int a, int b) { return _query(a, b, 0, 0, limit); }
};
RMQ rmq;
long long compute(long long cnt) {
  memset(fin, false, sizeof fin);
  for (int i = 0; i < N; ++i) {
    int L, R;
    L = i + N;
    R = i - cnt + N;
    if (L > R) swap(L, R);
    mini[i] = rmq.query(L, R + 1);
  }
  long long answer = x * cnt;
  deque<int> sp;
  for (int i = 0; i < N; ++i) {
    if (mini[i] == a[i]) {
      fin[i] = true;
      answer += a[i];
      sp.push_back(i);
    }
  }
  for (int i = 0; i < (int)sp.size(); ++i) {
    long long step = 0;
    int cur = (sp[i] + 1) % N;
    while (!fin[cur]) {
      (cur += 1) %= N;
      ++step;
      if (step > cnt) {
        answer += a[cur];
      } else {
        answer += a[sp[i]];
      }
    }
  }
  return answer;
}
int main() {
  cin >> N >> x;
  for (int i = 0; i < N; ++i) cin >> a[i];
  rmq.init(N * 2);
  {
    vector<int> vec;
    for (int i = 0; i < 2; ++i)
      for (int j = 0; j < N; ++j) vec.push_back(a[j]);
    rmq.build(vec);
  }
  long long mini = LLONG_MAX;
  for (int i = 0; i < N + 1; ++i) {
    mini = min(compute(i), mini);
  }
  cout << mini << endl;
  return 0;
}
