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;
using ll = long long;
using P = pair<long long, long long>;
long long dx[] = {1, 0, -1, 0};
long long dy[] = {0, 1, 0, -1};
constexpr ll mod = 1e9 + 7;
constexpr ll inf = INT32_MAX / 2;
constexpr ll INF = LLONG_MAX / 2;
constexpr long double eps = DBL_EPSILON;
constexpr long double pi = 3.141592653589793238462643383279;
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
ll fact(long long i) {
  if (i == 0) return 1;
  return (fact(i - 1)) * i % mod;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
long long keta(ll n) {
  if (n == 0) return 1;
  long long count = 0;
  while (n != 0) {
    n /= 10;
    count++;
  }
  return count;
}
ll ketasum(ll n) {
  ll sum = 0;
  while (n != 0) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
signed main() {
  long long n, x;
  cin >> n >> x;
  vector<long long> v(n);
  long long mi = inf;
  long long k;
  for (long long i = 0; i < (long long)(n); i++) {
    cin >> v[i];
    if (chmin(mi, v[i])) k = i;
  }
  long long minute = v[k];
  long long m = v[k] + x;
  for (long long i = k + 1; i < k + n; i++) {
    minute += min(v[i % n], m);
  }
  cout << minute << endl;
  return 0;
}
