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 MOD = 1e9 + 7;
int fx[] = {+1, -1, +0, +0, +1, +1, -1, -1, +0};
int fy[] = {+0, +0, +1, -1, +1, -1, +1, -1, +0};
template <typename T>
inline T GCD(T a, T b) {
  a = abs(a);
  b = abs(b);
  if (a < b) swap(a, b);
  while (b) {
    a = a % b;
    swap(a, b);
  }
  return a;
}
template <typename T>
inline T LCM(T x, T y) {
  T tp = GCD(x, y);
  if ((x / tp) * 1. * y > 9e18) return 9e18;
  return (x / tp) * y;
}
template <typename T>
inline T BigMod(T A, T B, T M = MOD) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % M;
    A = (A * A) % M;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M = MOD) {
  return BigMod(A, M - 2, M);
}
template <typename T>
T scani(T &n) {
  n = 0;
  bool negative = false;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') negative = true;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    n = n * 10 + c - 48;
    c = getchar();
  }
  if (negative) n = ~(n - 1);
  return n;
}
template <typename T>
void write(T n, int type = true) {
  if (n < 0) {
    putchar('-');
    n = -n;
  }
  if (!n) {
    putchar('0');
    if (type == 32)
      putchar(' ');
    else if (type)
      putchar('\n');
    return;
  }
  char buff[22];
  int len = 0;
  while (n) buff[len++] = n % 10 + 48, n /= 10;
  for (int i = len - 1; i >= 0; i--) putchar(buff[i]);
  if (type == 32)
    putchar(' ');
  else if (type)
    putchar('\n');
}
int scans(char *a) {
  int i = 0;
  char c = 0;
  while (c < 33) c = getchar();
  while (c > 33) {
    a[i++] = c;
    c = getchar();
  }
  a[i] = 0;
  return i;
}
const int N = 2222;
bool vis[N];
long long cost[N];
long long k;
int main() {
  int n;
  int rot = 0;
  long long sum = 0;
  scani(n), scani(k);
  for (int i = 0; i < n; i++) scani(cost[i]);
  for (int i = 0; i < n; i++) {
    if (cost[i] <= k) {
      vis[i] = true;
      sum += cost[i];
    }
  }
  while (true) {
    int mn = 0x3f3f3f3f;
    int id = -1;
    for (int j = 0; j < n; j++) {
      int pos = j - rot;
      if (pos < 0) pos += n;
      if (!vis[pos]) {
        if (mn > cost[j]) {
          mn = cost[j];
          id = j;
        }
      }
    }
    if (mn == 0x3f3f3f3f) {
      break;
    } else if (mn > k) {
      rot++;
      sum += k;
    } else {
      sum += mn;
      id = id - rot;
      if (id < 0) id += n;
      vis[id] = true;
    }
  }
  write(sum);
  return 0;
}
