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 double PI = acos(-1);
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
long long add(long long a, long long b) {
  long long ret = a + b;
  if (ret >= 1000000007) ret -= 1000000007;
  return ret;
}
long long subtract(long long a, long long b) {
  long long ret = a - b;
  if (ret < 0) ret += 1000000007;
  return ret;
}
long long mult(long long a, long long b) {
  long long ret = a * b;
  if (ret >= 1000000007) ret %= 1000000007;
  return ret;
}
long long bigmod(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = mult(ret, a);
    b >>= 1;
    a = mult(a, a);
  }
  return ret;
}
long long inverse(long long n) { return bigmod(n, 1000000007 - 2); }
bool base[1000010];
vector<int> primes;
void sieve(int mx) {
  mx += 10;
  int x = sqrt(mx);
  for (int i = 3; i <= x; i += 2)
    if (base[i] == 0)
      for (int j = i * i, k = i << 1; j < mx; j += k) base[j] = 1;
  primes.push_back(2);
  for (int i = 3; i < mx; i += 2)
    if (base[i] == 0) primes.push_back(i);
}
bool checkBit(int n, int i) { return (n & (1 << i)); }
int setBit(int n, int i) { return (n | (1 << i)); }
int resetBit(int n, int i) { return (n & (~(1 << i))); }
int n;
long long x, a[2010];
long long dp[2010][2010];
long long f(int i, int j) {
  if (i == j) return 0;
  long long &ret = dp[i][j];
  if (ret != -1) return ret;
  ret = a[i] + f((i + 1) % n, j);
  ret = min(ret, a[(j - 1 + n) % n] + f(i, (j - 1 + n) % n));
  ret = min(ret, x + f((i + 1) % n, (j + 1) % n));
  return ret;
}
int main() {
  scanf("%d %lld", &n, &x);
  for (int i = 0; i < n; ++i) scanf("%lld", &a[i]);
  memset(dp, -1, sizeof(dp));
  long long res = LLONG_MAX;
  for (int i = 0; i < n; ++i) res = min(res, a[i] + f((i + 1) % n, i));
  printf("%lld\n", res);
  return 0;
}
