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;
template <typename T>
inline void read(T &t) {
  t = 0;
  char c = getchar();
  long long f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -f;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    t = (t << 3) + (t << 1) + c - '0';
    c = getchar();
  }
  t *= f;
}
template <typename T, typename... Args>
inline void read(T &t, Args &...args) {
  read(t);
  read(args...);
}
template <typename T>
inline void write(T x) {
  if (x < 0) {
    x = -x;
    putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
long long n, x, val[2005], minn[2005][2005];
signed main() {
  read(n, x);
  for (register long long i = 1; i <= n; ++i) read(val[i]);
  for (register long long i = 1; i <= n; ++i) {
    minn[i][i] = val[i];
    for (register long long j = i + 1; j <= n; ++j)
      minn[i][j] = min(minn[i][j - 1], val[j]);
  }
  long long ans = 0x7f7f7f7f7f;
  for (register long long i = 0; i <= n; ++i) {
    long long res = 0;
    for (register long long j = 1; j <= n; ++j) {
      long long k = j - i;
      if (k <= 0)
        res += min(minn[n + k][n], minn[1][j]);
      else
        res += minn[k][j];
    }
    ans = min(ans, res + i * x);
  }
  write(ans), putchar('\n');
  return 0;
}
