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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
string toBin(long long a) {
  string res = "";
  while (a) {
    res += char((a & 1) + '0');
    a >>= 1;
  }
  reverse(res.begin(), res.end());
  return res;
}
const int mod = 1e9 + 7;
long long expo(long long base, long long pow) {
  long long ans = 1;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base *= base;
    base %= mod;
    pow /= 2;
  }
  return ans;
}
long long inv(long long x) { return expo(x, mod - 2); }
bool isPal(string ss) {
  int len = ss.length();
  for (int i = 0; i < len / 2; i++) {
    int comp = len - i - 1;
    if (ss[i] != ss[comp]) return false;
  }
  return true;
}
double pi = 3.141592653589793238462643;
double error = 0.0000001;
const int M = 10001;
int n;
long long x;
long long arr[M];
vector<long long> vec;
int main() {
  scanf("%d", &(n));
  scanf("%lld", &(x));
  int ind = -1;
  long long mi = mod;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &(arr[i]));
    if (arr[i] < mi) {
      mi = arr[i];
      ind = i;
    }
  }
  for (int i = ind; i < n; i++) {
    vec.push_back(arr[i]);
  }
  for (int i = 0; i < ind; i++) {
    vec.push_back(arr[i]);
  }
  int i = 0, j = 1;
  long long ans = 0;
  while (j < n) {
    if (vec[j] >= vec[i] + x) {
      j++;
    } else {
      int alp = j - i;
      ans += (alp * vec[i] + (alp - 1) * x);
      i = j;
      j++;
    }
  }
  int alp = j - i;
  ans += (alp * vec[i] + (alp - 1) * x);
  cout << ans;
}
