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>
const long long sqrtN = 64;
const long long INF = (1ll << 60ll);
long long n, x;
long long a[4096];
long long bucket[65];
void init(void);
long long range_min(int s, int t);
void update(int k, long long dat);
int main() {
  long long ans = INF;
  init();
  scanf("%lld %lld", &n, &x);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    update(i, a[i]);
    update(i + n, a[i]);
  }
  for (long long i = 0; i < n; i++) {
    long long a_sum = 0;
    for (long long j = 1; j <= n; j++) {
      long long k = j + n;
      a_sum += range_min(k - i, k + 1);
    }
    ans = ((ans) < (i * x + a_sum) ? (ans) : (i * x + a_sum));
  }
  printf("%lld\n", ans);
  return 0;
}
void init(void) {
  for (int i = 0; i < 4096; i++) {
    a[i] = INF;
  }
  for (int i = 0; i < 65; i++) {
    bucket[i] = INF;
  }
  return;
}
long long range_min(int s, int t) {
  long long minv = INF;
  long long K = (n + sqrtN - 1) / sqrtN;
  long long l, r;
  for (int k = 0; k < K; k++) {
    l = k * sqrtN;
    r = (k + 1) * sqrtN;
    if (t <= l) {
      break;
    } else if (r <= s) {
      continue;
    }
    if (s <= l && r <= t) {
      minv = ((minv) < (bucket[k]) ? (minv) : (bucket[k]));
    } else {
      for (long long i = ((l) > (s) ? (l) : (s)); i < ((r) < (t) ? (r) : (t));
           i++) {
        minv = ((minv) < (a[i]) ? (minv) : (a[i]));
      }
    }
  }
  return minv;
}
void update(int k, long long dat) {
  long long minv = INF;
  a[k] = dat;
  k /= sqrtN;
  for (long long i = k * sqrtN; i < (k + 1) * sqrtN; i++) {
    minv = ((minv) < (a[k]) ? (minv) : (a[k]));
  }
  bucket[k] = minv;
  return;
}
