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 int maxn = 2010;
int a[maxn * 2], n, x, r[maxn];
long long s[maxn * 2];
bool vis[maxn * 2];
int c[maxn * 2];
inline int lb(int x) { return x & -x; }
int read(int x) {
  int tot = 0;
  for (; x; x -= lb(x)) {
    tot += c[x];
  }
  return tot;
}
void add(int num, int x) {
  for (; x <= n + n; x += lb(x)) c[x] += num;
}
const int cmp(const int i, const int j) { return a[i] < a[j]; }
bool can(int x, int q) {
  int left, right;
  int l = 1, r = x;
  while (l < r) {
    int mid = (l + r) >> 1;
    if (read(x) - read(mid - 1) < x - mid + 1)
      l = mid + 1;
    else
      r = mid;
  }
  left = l;
  l = x, r = n + n;
  while (l < r) {
    int mid = (l + r + 1) >> 1;
    if (read(mid) - read(x - 1) < mid - x + 1)
      r = mid - 1;
    else
      l = mid;
  }
  right = l;
  return right - left + 1 <= q;
}
long long get_ans(int t) {
  for (int i = 1; i <= n; ++i) r[i] = i;
  memset(vis, 0, sizeof vis);
  memset(c, 0, sizeof c);
  sort(r + 1, r + 1 + n, cmp);
  for (int i = n; i >= 1; --i) {
    add(1, r[i]);
    add(1, n + r[i]);
    if (can(r[i], t) && can(n + r[i], t)) {
      vis[r[i]] = vis[n + r[i]] = 1;
    } else {
      add(-1, r[i]);
      add(-1, n + r[i]);
    }
  }
  long long ans = s[n] + 1ll * t * x;
  int b = 2;
  for (; b <= n + n; ++b) {
    if (vis[b] == 0 && vis[b - 1] == 1) break;
  }
  int p = b;
  for (int i = b; i <= b + n - 1; ++i) {
    if (vis[i]) ans -= a[i] - a[p];
    if (vis[i] == 0 && vis[i - 1]) p = b;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (int i = 1; i <= n + n; ++i) {
    if (i > n) a[i] = a[i - n];
    s[i] = s[i - 1] + a[i];
  }
  long long ans = s[n];
  for (int y = 0; y < n; ++y) {
    ans = min(ans, get_ans(y));
  }
  cout << ans << endl;
  return 0;
}
