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 int INF = 1000000007;
const int MAX_V = 201;
int now = 0;
const int MAX = 510000;
const int MOD = 1000000007;
long long fac[MAX], finv[MAX], inv[MAX];
void COMinit() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  for (int i = 2; i < MAX; i++) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}
long long COM(int n, int k) {
  if (n < k) return 0;
  if (n < 0 || k < 0) return 0;
  return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
struct edge {
  int t;
  edge(int _t) : t(_t) {}
};
bool use[100010];
vector<edge> te[100010];
vector<edge> fe[100010];
int t[64];
int d[100100];
long long rec(int num) {
  if (num == 0) return 0;
  return rec(num - 1) + num;
}
void Solve() {
  int n;
  long long x;
  cin >> n >> x;
  long long a[2010];
  long long best = 0;
  for (size_t i = 0; i < n; i++) {
    cin >> a[i];
    best += a[i];
  }
  long long aval[2010][2010];
  for (size_t i = 0; i < n; i++) {
    aval[0][i] = a[i];
  }
  long long *prev = aval[0];
  long long *next = aval[1];
  int maxn = 0;
  for (int i = 1; i < n; i++) {
    int count = 0;
    bool find = false;
    long long atem[2010];
    for (int j = 0; j < n; j++) {
      aval[i][j] = aval[i - 1][j];
      atem[j] = aval[i - 1][j];
    }
    long long su = 0;
    for (int j = 0; j < n; j++) {
      int from = (j + n - i) % n;
      if (aval[i][j] > a[from] + x * i) {
        find = true;
        atem[j] = a[from];
      }
      if (aval[i][j] > a[from]) {
        atem[j] = a[from];
        count++;
      }
      su += atem[j];
    }
    if (su + i * x < best) {
      best = su + i * x;
    } else
      continue;
    if (find || count > 1) {
      maxn = i;
      for (size_t j = 0; j < n; j++) {
        aval[i][j] = atem[j];
      }
    }
  }
  long long sum = 0;
  for (size_t i = 0; i < n; i++) {
    sum += aval[n - 1][i];
  }
  cout << (sum + maxn * x) << endl;
}
int main() {
  Solve();
  return 0;
}
