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;
bool prime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return n != 1;
}
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long lcm(long long x, long long y) { return x * y / gcd(x, y); }
long long mod_pow(long long x, long long y, long long mod) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) {
      res = res * x % mod;
    }
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
long long kai(long long x) {
  if (x == 0) return 1;
  return (kai(x - 1) * x) % 998244353;
}
long long comb(long long x, long long y) {
  return kai(x) * mod_pow(kai(x - y), 998244353 - 2, 998244353) % 998244353 *
         mod_pow(kai(y), 998244353 - 2, 998244353) % 998244353;
}
struct edge {
  long long to, cost;
};
long long dx[5] = {0, 1, 0, -1}, dy[5] = {1, 0, -1, 0};
map<long long, long long> factor(long long x) {
  map<long long, long long> res;
  for (long long i = 2; i * i <= x; i++) {
    while (x % i == 0) {
      x /= i;
      res[i]++;
    }
  }
  if (x != 1) res[x]++;
  return res;
}
pair<long long, long long> ans[2019];
long long n, x, a[2019];
signed main() {
  cin >> n >> x;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) {
    long long memo = 5000000000000000000;
    for (long long j = 0; j < n; j++) {
      if (i >= j) {
        if (memo > a[j] + x * (i - j)) {
          ans[i].first = a[j];
          ans[i].second = x * (i - j);
          memo = a[j] + x * (i - j);
        }
      } else {
        if (memo > a[j] + x * (n - j + i)) {
          ans[i].first = a[j];
          ans[i].second = x * (n - j + i);
          memo = a[j] + x * (n - j + i);
        }
      }
    }
  }
  long long res = 0, ma = 0;
  for (long long i = 0; i < n; i++) {
    res += ans[i].first;
    ma = max(ma, ans[i].second);
  }
  cout << res + ma << endl;
  getchar();
  getchar();
  return 0;
}
