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;
long long N, x;
long long a[2005];
long long w[2005];
struct segtree {
  vector<long long> data;
  long long n;
  segtree(long long nin) {
    n = pow(2, ceil(log(nin) / log(2)));
    long long size = 2 * n - 1;
    this->data.resize(size, 1e9);
  }
  void update(long long idx, long long val) { update(idx, val, 0, 0, n); }
  long long update(long long idx, long long val, long long node_id,
                   long long left, long long right) {
    if (!(left <= idx && idx < right)) {
      return data[node_id];
    }
    if (left + 1 < right) {
      long long mid = (left + right) / 2;
      long long L = update(idx, val, 2 * node_id + 1, left, mid);
      long long R = update(idx, val, 2 * node_id + 2, mid, right);
      data[node_id] = min(L, R);
    } else {
      data[node_id] = min(data[node_id], val);
    }
    return data[node_id];
  }
  long long get(long long l, long long r) { return get(l, r, 0, 0, n); }
  long long get(long long query_l, long long query_r, long long node_id,
                long long node_l, long long node_r) {
    if (query_r <= node_l || node_r <= query_l) {
      return INT_MAX / 3;
    }
    if (query_l <= node_l && node_r <= query_r) {
      return data[node_id];
    }
    long long mid = (node_l + node_r) / 2;
    long long L = get(query_l, query_r, 2 * node_id + 1, node_l, mid);
    long long R = get(query_l, query_r, 2 * node_id + 2, mid, node_r);
    return min(L, R);
  }
};
long long cal(int slide) {
  for (int i = 0; i < N; i++) {
    w[i] = a[i];
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j <= slide; j++) {
      w[(i + j) % N] = min(w[(i + j) % N], a[i]);
    }
  }
  long long ret = 0;
  for (int i = 0; i < N; i++) {
    ret += w[i];
  }
  return ret + slide * x;
}
int main() {
  cin >> N >> x;
  segtree sg(N);
  for (int i = 0; i < N; i++) {
    cin >> a[i];
    sg.update(i, a[i]);
  }
  long long ret = INT_MAX;
  for (int slide = 0; slide < N; slide++) {
    long long s = 0;
    for (int i = 0; i < N; i++) {
      long long tmp = a[i];
      long long left = i - slide;
      long long right = i;
      if (left < 0) {
        tmp = min(tmp, sg.get((left + N) % N, N));
      }
      tmp = min(tmp, sg.get(max((long long)0, left), right));
      s += tmp;
    }
    ret = min(ret, s + slide * x);
  }
  cout << ret << endl;
}
