Vasya takes part in the orienteering competition. There are n checkpoints located along the line at coordinates x1, x2, ..., xn. Vasya starts at the point with coordinate a. His goal is to visit at least n - 1 checkpoint in order to finish the competition. Participant are allowed to visit checkpoints in arbitrary order.

Vasya wants to pick such checkpoints and the order of visiting them that the total distance travelled is minimized. He asks you to calculate this minimum possible value.

Input

The first line of the input contains two integers n and a (1 ≤ n ≤ 100 000,  - 1 000 000 ≤ a ≤ 1 000 000) — the number of checkpoints and Vasya's starting position respectively.

The second line contains n integers x1, x2, ..., xn ( - 1 000 000 ≤ xi ≤ 1 000 000) — coordinates of the checkpoints.

Output

Print one integer — the minimum distance Vasya has to travel in order to visit at least n - 1 checkpoint.

Examples

Input

3 10
1 7 12


Output

7


Input

2 0
11 -10


Output

10


Input

5 0
0 0 1000 0 0


Output

0

Note

In the first sample Vasya has to visit at least two checkpoints. The optimal way to achieve this is the walk to the third checkpoints (distance is 12 - 10 = 2) and then proceed to the second one (distance is 12 - 7 = 5). The total distance is equal to 2 + 5 = 7.

In the second sample it's enough to visit only one checkpoint so Vasya should just walk to the point  - 10.
#include <bits/stdc++.h>
using namespace std;
int inf_int = 2e9;
long long inf_ll = 2e18;
const double pi = 3.1415926535898;
template <typename T, typename T1>
void prin(vector<pair<T, T1> >& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i].first << " " << a[i].second << "\n";
  }
}
template <typename T, typename T1>
void prin(set<pair<T, T1> >& a) {
  for (auto it = a.begin(); it != a.end(); it++) {
    cout << it->first << " " << it->second << "\n";
  }
}
template <typename T>
void prin(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i];
    if (i < a.size() - 1)
      cout << " ";
    else
      cout << "\n";
  }
}
template <typename T>
void prin(set<T>& a) {
  for (auto it = a.begin(); it != a.end(); it++) {
    cout << *it << " ";
  }
}
template <typename T>
void prin_new_line(vector<T>& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << "\n";
  }
}
template <typename T, typename T1>
void prin_new_line(vector<pair<T, T1> >& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i].first << " " << a[i].second << "\n";
  }
}
int sum_vec(vector<int>& a) {
  int s = 0;
  for (int i = 0; i < a.size(); i++) {
    s += a[i];
  }
  return s;
}
template <typename T>
T max(vector<T>& a) {
  T ans = a[0];
  for (int i = 1; i < a.size(); i++) {
    ans = max(ans, a[i]);
  }
  return ans;
}
template <typename T>
T min(vector<T>& a) {
  T ans = a[0];
  for (int i = 1; i < a.size(); i++) {
    ans = min(ans, a[i]);
  }
  return ans;
}
template <typename T>
T min(T a, T b, T c) {
  return min(a, min(b, c));
}
template <typename T>
T max(T a, T b, T c) {
  return max(a, max(b, c));
}
long double s_triangle(long double x1, long double y1, long double x2,
                       long double y2, long double x3, long double y3) {
  return abs(((x2 - x1) * (y3 - y1) - (x3 - x1) * (y2 - y1)) / 2);
}
bool overflow(long long a, long long b) {
  if (a * b / b != a) return true;
  return false;
}
bool debug = 0;
const int maxn = 1e5 + 100;
long long mod = 1e9 + 7;
void solve() {
  long long n, a;
  cin >> n >> a;
  vector<long long> l, r;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (x >= a) {
      r.push_back(x);
    } else {
      l.push_back(x);
    }
  }
  sort(r.begin(), r.end());
  sort(l.begin(), l.end());
  reverse(l.begin(), l.end());
  if (n == 1) {
    cout << 0;
    return;
  }
  int ans = inf_int;
  for (int i = 0; i <= r.size(); i++) {
    int x = n - 1 - i;
    if (x <= l.size()) {
      int gfgs, x2;
      if (i > 0) {
        gfgs = r[i - 1] - a;
      } else {
        gfgs = 0;
      }
      if (x > 0) {
        x2 = a - l[x - 1];
      } else {
        x2 = 0;
      }
      ans = min(ans, 2 * gfgs + x2, 2 * x2 + gfgs);
    }
  }
  cout << ans;
}
int main() {
  if (!debug) {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  int t = 1;
  while (t--) solve();
  return 0;
}
