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 int INF = 1001001001;
const long long LINF = 1e17;
const double pi = 3.1415926535897932;
const string endstr = "\n";
template <typename T>
T gcd(T a, T b) {
  return (a == 0) ? b : gcd(b % a, a);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
bool p_comp_fs(const pair<long long, long long> p1,
               const pair<long long, long long> p2) {
  return p1.first < p2.first;
};
bool p_comp_fg(const pair<long long, long long> p1,
               const pair<long long, long long> p2) {
  return p1.first > p2.first;
};
bool p_comp_ss(const pair<long long, long long> p1,
               const pair<long long, long long> p2) {
  return p1.second < p2.second;
};
bool p_comp_sg(const pair<long long, long long> p1,
               const pair<long long, long long> p2) {
  return p1.second > p2.second;
};
template <typename T>
vector<T> uniquen(vector<T> vec) {
  sort((vec).begin(), (vec).end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  return vec;
}
inline long long popcnt(long long x) {
  return __builtin_popcountll((unsigned long long)x);
};
template <class T>
bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
long long N, X;
vector<long long> A;
long long cost(long long K) {
  long long res = K * X;
  vector<long long> mns(N, LINF);
  for (long long i = 0; i < N; i++) {
    long long mn = LINF;
    for (long long j = 0; j < K + 1; j++) {
      chmin(mn, A[(i - j + N) % N]);
    }
    mns[i] = mn;
  }
  for (long long i = 0; i < N; i++) res += mns[i];
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> N >> X;
  A = vector<long long>(N);
  for (long long i = 0; i < N; i++) cin >> A[i];
  long long ans = LINF;
  for (long long k = 0; k < N; k++) chmin(ans, cost(k));
  cout << ans << endstr;
  return 0;
}
