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;
int a[2000];
struct pqElem {
  int index;
  long long sum;
  pqElem(int index, long long sum) {
    this->index = index;
    this->sum = sum;
  }
  bool operator<(const pqElem &other) const {
    if (sum == other.sum)
      return index < other.index;
    else
      return sum > other.sum;
  }
};
int main() {
  int n, x;
  scanf("%d %d", &n, &x);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int index = min_element(a, a + n) - a;
  for (int p = 0; p < index; p++) {
    int temp = a[0];
    for (int i = 1; i < n; i++) a[i - 1] = a[i];
    a[n - 1] = temp;
  }
  long long ans = (1ll << 62);
  for (int p = 0; p < n; p++) {
    priority_queue<pqElem> pq;
    for (int i = n - 1, j = 0; j <= p - 1; j++, i--)
      pq.push(pqElem(i, a[i] + 1ll * (n - 1 - i) * x));
    long long cost = 0;
    for (int i = n - 1; i >= 0; i--) {
      pq.push(pqElem(i - p, a[i - p] + 1ll * (n - 1 - (i - p)) * x));
      while (pq.top().index > i) pq.pop();
      cost += a[pq.top().index];
    }
    cost += x * 1ll * p;
    ans = min(ans, cost);
  }
  printf("%lld\n", ans);
  return 0;
}
