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;
typedef unsigned long long ullong;
struct A {
  ullong n;
  A *next, *prev;
};
vector<ullong> first_search(A *aaa, ullong n, ullong x) {
  vector<ullong> v;
  for (ullong i = 0; i < n; i++) {
    if (aaa[i].n < x) v.push_back(i);
  }
  return v;
}
int main() {
  ullong time = 0;
  ullong n, x;
  cin >> n >> x;
  A *a = new A[n];
  a[0].prev = &a[n - 1];
  a[n - 1].next = &a[0];
  for (ullong i = 0; i < n - 2; i++) a[i].next = &a[i + 1];
  for (ullong i = 1; i < n - 1; i++) a[i].prev = &a[i - 1];
  for (ullong i = 0; i < n; i++) cin >> a[i].n;
  vector<ullong> first = first_search(a, n, x);
  bool flag;
  cout << "first.size()=" << first.size() << endl;
  for (ullong i = 0; i < n; i++) {
    flag = false;
    for (ullong j = 0; j < first.size(); j++) {
      if (i == first[j]) flag = true;
    }
    if (flag) {
      time += a[i].n;
    } else {
      time += x;
    }
  }
  cout << time << endl;
}
