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>
#pragma comment(linker, "/STACK:1024000000")
using namespace std;
void solve(bool);
void precalc();
clock_t start;
int testNumber = 1;
bool todo = true;
bool stress = false;
int main() {
  start = clock();
  int t = 1;
  cout.sync_with_stdio(0);
  cin.tie(0);
  precalc();
  cout.precision(10);
  cout << fixed;
  int testNum = 1;
  while (t--) {
    solve(true);
    ++testNumber;
  }
  return 0;
}
template <typename T>
T binpow(T q, T w, T mod) {
  if (!w) return 1 % mod;
  if (w & 1) return q * 1LL * binpow(q, w - 1, mod) % mod;
  return binpow(q * 1LL * q % mod, w / 2, mod);
}
void precalc() {}
template <typename T>
T gcd(T q, T w) {
  while (w) {
    q %= w;
    swap(q, w);
  }
  return q;
}
template <typename T>
T lcm(T q, T w) {
  return q / gcd(q, w) * w;
}
void solve(bool read) {
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
  }
  long long ans = 1e18;
  for (long long use = 0; use < n; ++use) {
    long long res = use * x;
    for (long long i = 0; i < n; ++i) {
      long long mn = a[i];
      for (long long j = i, cnt = 0; cnt <= use; j = (j - 1 + n) % n, ++cnt) {
        mn = min(mn, a[j]);
      }
      res += mn;
    }
    ans = min(ans, res);
  }
  cout << ans << "\n";
}
