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;
long long nextInt() {
  long long a;
  cin >> a;
  return a;
}
char nextChar() {
  char a;
  cin >> a;
  return a;
}
double nextDouble() {
  double a;
  cin >> a;
  return a;
}
string nextString() {
  string a;
  cin >> a;
  return a;
}
void inputVector(vector<long long> &v, long long &n) {
  for (long long i = 0; i < (n); i++) {
    v.push_back(nextInt());
  }
}
void inputVector(vector<double> &v, long long &n) {
  for (long long i = 0; i < (n); i++) {
    v.push_back(nextDouble());
  }
}
void inputVector(vector<string> &v, long long &n) {
  for (long long i = 0; i < (n); i++) {
    v.push_back(nextString());
  }
}
long long N;
long long dist(long long a, long long b) {
  if (a <= b) return b - a;
  return a + N + 1 - b;
}
signed main() {
  long long x;
  cin >> N >> x;
  vector<long long> a;
  inputVector(a, N);
  vector<long long> cnt(N, 0);
  for (long long i = 0; i < (N); i++) {
    long long minCost = LLONG_MAX;
    long long minArg = -1;
    for (long long j = 0; j < (N); j++) {
      long long cost = a[j] + dist(i, j) * x;
      if (cost < minCost) {
        minCost = cost;
        minArg = j;
      }
    }
    cnt[minArg]++;
  }
  long long cost = 0;
  long long maxCnt = 0;
  for (long long i = 0; i < (N); i++) {
    maxCnt = max(maxCnt, cnt[i]);
    cost += cnt[i] * a[i];
  }
  cout << cost + (maxCnt - 1) * x << endl;
  return 0;
}
