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 warning(disable : 4244)
#pragma GCC optimize(2)
using namespace std;
const long long ll_inf = 9223372036854775807;
const int int_inf = 2147483647;
const short short_inf = 32767;
const char char_inf = 127;
long long GCD(long long a, long long b) { return a ? GCD(b, a % b) : a; }
inline long long read() {
  long long c = getchar(), Nig = 1, x = 0;
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') Nig = -1, c = getchar();
  while (isdigit(c)) x = ((x << 1) + (x << 3)) + (c ^ '0'), c = getchar();
  return Nig * x;
}
inline void out(long long a) {
  if (a < 0) putchar('-'), a = -a;
  if (a >= 10) out(a / 10);
  putchar(a % 10 + '0');
}
long long qpow(long long x, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    n >>= 1;
  }
  return res;
}
int save[20000];
void S(int x, int n);
int dp[3000][3000];
int main() {
  int n = read(), x = read();
  for (int i = 1; i <= n; i++) save[i] = read();
  for (int i = 1; i <= n; i++) {
    dp[i][0] = save[i];
    for (int j = 1; j < n; j++) {
      int temp = i - j;
      if (temp > 0)
        ;
      else
        temp += n;
      dp[i][j] = min(dp[i][j - 1], save[temp]);
    }
  }
  S(x, n);
}
void S(int x, int n) {
  int ans = 0;
  int res = int_inf;
  for (int i = 0; i < n; i++) {
    ans = x * i;
    for (int j = 1; j <= n; j++) ans += dp[j][i];
    res = min(ans, res);
  }
  cout << res << endl;
}
