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 long INF = 1e18;
const long long MOD = 1e9 + 7;
const double pi = acos(-1);
const int SIZE = 1 << 17;
int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1}, alp[30];
long long fac[200005], finv[200005], inv[200005];
vector<long long> dij;
struct edge {
  long long to, cost;
};
vector<vector<edge> > G;
long long mod_pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res = res * a % MOD;
    }
    a = a * a % MOD;
    b >>= 1;
  }
  return res;
}
void addedge(int from, int to, int cost) {
  G[from].push_back({to, cost});
  G[to].push_back({from, cost});
}
long long a, b, k[2005], l[2005], ans = INF, mon;
int main() {
  cin >> a >> b;
  for (int i = 0; i < a; ++i) {
    cin >> k[i];
    l[i] = k[i];
  }
  for (int i = 1; i < a; ++i) {
    mon = i * b;
    for (int j = 0; j < a; ++j) {
      l[j] = min(l[j], k[(a + j - i) % a]);
      mon += l[j];
    }
    ans = min(ans, mon);
  }
  cout << ans << endl;
}
