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;
const long double EPS = 1e-9;
const long double PI = acos(-1.0);
const int INF = (int)1e9 + 5;
const long long LINF = (long long)1e18 + 5;
bool eq(double a, double b) { return fabs(a - b) < EPS; }
template <class T>
T sqr(const T& x) {
  return x * x;
}
template <class T>
long long abs(const T& x) {
  return x < 0 ? -x : x;
}
template <class T>
long long myround(const T& x) {
  return x < 0 ? x - 0.5 : x + 0.5;
}
template <class T>
bool chmin(T& x, const T& y) {
  if (y < x) {
    x = y;
    return true;
  }
  return false;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  os << "{";
  for (typename vector<T>::const_iterator it = v.begin(); it != v.end(); ++it) {
    os << *it;
    if ((it + 1) != v.end()) os << ", ";
  }
  os << "}";
  return os;
}
template <class P, class Q>
ostream& operator<<(ostream& os, const pair<P, Q>& p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
bool chmax(T& x, const T& y) {
  if (x < y) {
    x = y;
    return true;
  }
  return false;
}
const int N = 2020;
int n, x;
int startCost[N], cost[N], oldCost[N];
int main() {
  srand(time(0));
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> startCost[i];
    cost[i] = startCost[i];
  }
  long long res = LINF;
  for (int s = 0; s < n; s++) {
    long long cur = 1LL * s * x;
    for (int i = 0; i < n; i++) {
      cur += cost[i];
    }
    res = min(res, cur);
    memcpy(oldCost, cost, sizeof(cost[0]) * n);
    for (int j = 0; j < n; j++) {
      cost[j] = min(oldCost[j], oldCost[(j + s + 1) % n]);
    }
  }
  cout << res << endl;
}
