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;
std::clock_t __start;
double __duration;
void start_timer() { __start = std::clock(); }
void print_timer() {
  __duration = (std::clock() - __start) / (double)CLOCKS_PER_SEC;
  std::cout << "Duration (msec): " << __duration * 1000 << '\n';
}
template <class T>
T min_(T a, T b) {
  return (a < b ? a : b);
}
template <class T>
T max_(T a, T b) {
  return (a > b ? a : b);
}
double EPS = 1e-9;
bool eq_(const double& lhs, const double& rhs) {
  return (fabs(lhs - rhs) < EPS);
}
const int INF = int(2e9);
const long long INF_LL = LLONG_MAX;
vector<int> prices;
int N, X;
vector<vector<int> > DP;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> X;
  prices.resize(N);
  for (int i = 0; i < N; ++i) {
    cin >> prices[i];
  }
  DP.assign(N + 1, vector<int>(N + 1, 0));
  for (int i = 0; i < N; ++i) {
    DP[i][i] = prices[i];
  }
  for (int len = 1; len < N; ++len) {
    for (int L = 0; L + len < N; ++L) {
      int R = L + len;
      DP[L][R] =
          min(DP[L][R - 1] + prices[R],
              min(prices[L] + DP[L + 1][R], DP[L + 1][R] + prices[R] + X));
    }
  }
  cout << DP[0][N - 1] << '\n';
}
