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 <class T>
inline T BM(T p, T e, T M) {
  long long int ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T mdINV(T a, T M) {
  return BM(a, M - 2, M);
}
template <class T>
inline T PW(T p, T e) {
  long long int ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p);
    p = (p * p);
  }
  return (T)ret;
}
template <class T>
string NTS(T Number) {
  stringstream ss;
  ss << Number;
  return ss.str();
}
template <class T>
T stringtonumber(const string &Text) {
  istringstream ss(Text);
  T result;
  return ss >> result ? result : 0;
}
template <class T>
bool ISLEFT(T a, T b, T c) {
  if (((a.first.first - b.first.first) * (b.second.second - c.second.second) -
       (b.first.first - c.first.first) * (a.second.second - b.second.second)) <
      0.0)
    return 1;
  else
    return 0;
}
int main() {
  long long int n, m;
  scanf("%lld%lld", &n, &m);
  long long int res = 0;
  long long int ans = 2e9;
  for (int i = 1; i <= n; i++) {
    long long int first;
    scanf("%lld", &first);
    ans = min(ans, first);
    res += first;
  }
  ans = min(res, n + (n - 1) * m);
  printf("%lld\n", ans);
  return 0;
}
