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;
double e_time(void) {
  static struct timeval now;
  gettimeofday(&now, NULL);
  return (double)(now.tv_sec + now.tv_usec / 1000000.0);
}
template <class T>
void vp(T& a, int p) {
  for (int i = 0; i < p; i++) cout << a[i] << " ";
  cout << endl;
}
template <class T>
T ceilUp(const T& a, const T& b) {
  return (a + b - 1) / b;
}
const long long mod = 1000000009;
inline long long modadd(long long a, long long b) {
  a %= mod;
  b %= mod;
  return (a + b) % mod;
}
inline long long modsub(long long a, long long b) {
  a %= mod;
  b %= mod;
  a -= b;
  a %= mod;
  if (a < 0) a += mod;
  return a;
}
inline long long modmul(long long a, long long b) {
  a %= mod;
  b %= mod;
  return (a * b) % mod;
}
inline long long modeq(long long a, long long b, long long m) {
  a = (a % m + m) % m;
  b = (b % m + m) % m;
  return a == b;
}
long long mypow(long long n, long long p) {
  if (p == 0) return 1;
  if (p == 1) return n;
  long long ret = mypow(n, p / 2);
  ret = ret * ret;
  if (p & 1) ret = ret * n;
  return ret;
}
const int N = 2000;
int table[N][N];
long long solve(int n, vector<int>& in, int x) {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      table[i][j] = in[(i - j + n) % n];
    }
    for (int j = 1; j < n; j++) {
      table[i][j] = min(table[i][j], table[i][j - 1]);
    }
  }
  long long ret = accumulate(begin(in), end(in), 0LL);
  for (int j = 0; j < n; j++) {
    long long tmp = j * x;
    for (int i = 0; i < n; i++) {
      tmp += table[i][j];
    }
    ret = min(ret, tmp);
  }
  return ret;
}
int main() {
  int n, x;
  cin >> n >> x;
  vector<int> in(n);
  for (int i = 0; i < n; i++) cin >> in[i];
  cout << solve(n, in, x) << endl;
  return 0;
}
