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 N = 2e3 + 3;
const long long inf = 1e15;
int n, x;
int arr[N];
long long dist[N][N];
priority_queue<pair<long long, pair<int, int> > > pq;
void relax(int l, int r, long long cost) {
  if (dist[l][r] > cost) {
    dist[l][r] = cost;
    pq.push({-cost, {l, r}});
  }
}
int main() {
  scanf("%d %d", &n, &x);
  for (int i = 0; i < n; ++i) {
    scanf("%d", arr + i);
  }
  while (!pq.empty()) {
    pq.pop();
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      dist[i][j] = inf;
    }
  }
  for (int i = 0; i < n; ++i) {
    dist[i][i] = arr[i];
    pq.push({-arr[i], {i, i}});
  }
  long long ans = inf;
  while (!pq.empty()) {
    int l = pq.top().second.first;
    int r = pq.top().second.second;
    long long cost = -pq.top().first;
    pq.pop();
    if ((r + 1) % n == l) {
      ans = min(ans, cost);
    }
    if (cost > dist[l][r]) {
      continue;
    }
    relax((l + 1) % n, (r + 1) % n, cost + x);
    relax((l + n - 1) % n, r, cost + arr[(l + n - 1) % n]);
    relax(l, (r + 1) % n, cost + arr[(r + 1) % n]);
  }
  printf("%lld\n", ans);
}
