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;
}
struct Eflag {
  bool flag;
  Eflag *prev;
};
bool chech(ullong k, vector<ullong> first) {
  for (ullong i = 0; i < first.size(); i++) {
    if (k == first[i]) return true;
  }
  return false;
}
ullong length(A, ullong);
int main() {
  ullong n, x;
  cin >> n >> x;
  ullong time = 0;
  A *a = new A[n];
  a[0].prev = &a[n - 1];
  a[n - 1].next = &a[0];
  for (ullong i = 0; i < n - 1; i++) a[i].next = &a[i + 1];
  for (ullong i = 1; i < n; i++) a[i].prev = &a[i - 1];
  for (ullong i = 0; i < n; i++) cin >> a[i].n;
  a[0];
  vector<ullong> first = first_search(a, n, x);
  A *f = new A[first.size()];
  if (first.size() >= 2) {
    f[0].prev = &f[first.size() - 1];
    f[first.size() - 1].next = &f[0];
    for (ullong i = 0; i < first.size() - 1; i++) f[i].next = &f[i + 1];
    for (ullong i = 1; i < first.size(); i++) f[i].prev = &f[i - 1];
    for (ullong i = 0; i < first.size(); i++) {
      f[i].n = first[i];
    }
  } else if (first.size() == 1) {
    f[0].next = &f[0];
    f[0].n = first[0];
  }
  ullong max = 0;
  if (first.size() >= 2) {
    for (ullong i = 0; i < first.size(); i++) {
      ullong work;
      work = length(a[f[i].n], a[f[i].next->n].n);
      if (work > max) max = work;
      time += (work * a[f[i].n].n);
    }
    time += ((max - 1) * x);
  } else if (first.size() == 1) {
    time += (n * a[f[0].n].n);
    time += (x * (n - 1));
  } else {
    for (ullong i = 0; i < n; i++) {
      time += a[i].n;
    }
  }
  cout << time << endl;
}
ullong length(A a, ullong i) {
  if (a.n != i)
    return 1 + length(*a.next, i);
  else
    return 0;
}
