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 long long mod = 1000000007;
vector<long long> a;
int N;
long long X;
long long get(int x) {
  long long res = x * X;
  deque<long long> mono;
  int j = 0, i = 0, st = (N - x) % N;
  for (j = N - x; j < N; j++) {
    while (!mono.empty() && mono.back() > a[j]) mono.pop_back();
    mono.push_back(a[j]);
  }
  for (i = 0; i < N; i++) {
    while (!mono.empty() && mono.back() > a[i]) mono.pop_back();
    mono.push_back(a[i]);
    assert(!mono.empty());
    res += mono.front();
    if (!mono.empty() && mono.front() == a[st]) mono.pop_back();
    st++;
    if (st > N) st -= N;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> N >> X;
  a.resize(N);
  long long res = LLONG_MAX;
  for (int i = 0; i < N; i++) cin >> a[i];
  for (int i = 0; i < N; i++) res = min(res, get(i));
  cout << res << '\n';
  return 0;
}
