#include <bits/stdc++.h>
long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  long long x;
  long long y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  long long x;
  long long y, z, w;
  pa4(long long x = 0, long long y = 0, long long z = 0, long long w = 0)
      : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  long long x, y;
  pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long gcd(long long v, long long b) {
  if (v == 0) return b;
  if (b == 0) return v;
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
long long mod;
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<long long, long long> operator+(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first + r.first, l.second + r.second};
}
pair<long long, long long> operator-(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first - r.first, l.second - r.second};
}
long long pr[1000010];
long long inv[1000010];
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
long long comb(long long nn, long long rr) {
  if (rr < 0 || rr > nn || nn < 0) return 0;
  long long r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(long long ert) {
  pr[0] = 1;
  for (long long i = 1; i <= ert; i++) {
    pr[i] = (pr[i - 1] * i) % mod;
  }
  inv[ert] = beki(pr[ert], mod - 2, mod);
  for (long long i = ert - 1; i >= 0; i--) {
    inv[i] = inv[i + 1] * (i + 1) % mod;
  }
}
long long a[100020][3];
vector<long long> ve;
vector<long long> G[100020];
void dfs(long long r, long long p) {
  ve.push_back(r);
  for (auto v : G[r])
    if (v != p) dfs(v, r);
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  map<long long, long long> ma;
  for (long long i = 0; i < n; i++) {
    long long y;
    cin >> y;
    ma[y]++;
  }
  deque<pair<long long, long long> > de;
  for (auto it = ma.begin(); it != ma.end(); it++) {
    pair<long long, long long> z = *it;
    de.push_back(z);
  }
  while (de.size() > 1) {
    n = de.size();
    pair<long long, long long> a = de.front();
    pair<long long, long long> b = de.back();
    if (k == 0) break;
    if (a.second < b.second) {
      long long sa = de[1].first - de[0].first;
      if (k < a.second * sa) {
        de[0].first += k / a.second;
        break;
      } else {
        k -= a.second * sa;
        de.pop_front();
        de[0].second += a.second;
      }
    } else {
      long long sa = de[n - 1].first - de[n - 2].first;
      if (k < b.second * sa) {
        de[n - 1].first -= k / b.second;
        break;
      } else {
        k -= b.second * sa;
        de.pop_back();
        de.back().second += b.second;
      }
    }
  }
  cout << -de[0].first + de.back().first << endl;
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

You are given a sequence a_1, a_2, ..., a_n consisting of n integers.

You may perform the following operation on this sequence: choose any element and either increase or decrease it by one.

Calculate the minimum possible difference between the maximum element and the minimum element in the sequence, if you can perform the aforementioned operation no more than k times.

Input

The first line contains two integers n and k (2 ≤ n ≤ 10^{5}, 1 ≤ k ≤ 10^{14}) — the number of elements in the sequence and the maximum number of times you can perform the operation, respectively.

The second line contains a sequence of integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^{9}).

Output

Print the minimum possible difference between the maximum element and the minimum element in the sequence, if you can perform the aforementioned operation no more than k times.

Examples

Input


4 5
3 1 7 5


Output


2


Input


3 10
100 100 100


Output


0


Input


10 9
4 5 5 7 5 4 5 2 4 3


Output


1

Note

In the first example you can increase the first element twice and decrease the third element twice, so the sequence becomes [3, 3, 5, 5], and the difference between maximum and minimum is 2. You still can perform one operation after that, but it's useless since you can't make the answer less than 2.

In the second example all elements are already equal, so you may get 0 as the answer even without applying any operations.

### Response

```cpp
#include <bits/stdc++.h>
long long dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
long long dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
using namespace std;
class pa3 {
 public:
  long long x;
  long long y, z;
  pa3(long long x = 0, long long y = 0, long long z = 0) : x(x), y(y), z(z) {}
  bool operator<(const pa3 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    return z < p.z;
  }
  bool operator>(const pa3 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    return z > p.z;
  }
  bool operator==(const pa3 &p) const {
    return x == p.x && y == p.y && z == p.z;
  }
  bool operator!=(const pa3 &p) const {
    return !(x == p.x && y == p.y && z == p.z);
  }
};
class pa4 {
 public:
  long long x;
  long long y, z, w;
  pa4(long long x = 0, long long y = 0, long long z = 0, long long w = 0)
      : x(x), y(y), z(z), w(w) {}
  bool operator<(const pa4 &p) const {
    if (x != p.x) return x < p.x;
    if (y != p.y) return y < p.y;
    if (z != p.z) return z < p.z;
    return w < p.w;
  }
  bool operator>(const pa4 &p) const {
    if (x != p.x) return x > p.x;
    if (y != p.y) return y > p.y;
    if (z != p.z) return z > p.z;
    return w > p.w;
  }
  bool operator==(const pa4 &p) const {
    return x == p.x && y == p.y && z == p.z && w == p.w;
  }
};
class pa2 {
 public:
  long long x, y;
  pa2(long long x = 0, long long y = 0) : x(x), y(y) {}
  pa2 operator+(pa2 p) { return pa2(x + p.x, y + p.y); }
  pa2 operator-(pa2 p) { return pa2(x - p.x, y - p.y); }
  bool operator<(const pa2 &p) const { return y != p.y ? y < p.y : x < p.x; }
  bool operator>(const pa2 &p) const { return x != p.x ? x < p.x : y < p.y; }
  bool operator==(const pa2 &p) const {
    return abs(x - p.x) == 0 && abs(y - p.y) == 0;
  }
  bool operator!=(const pa2 &p) const {
    return !(abs(x - p.x) == 0 && abs(y - p.y) == 0);
  }
};
string itos(long long i) {
  ostringstream s;
  s << i;
  return s.str();
}
long long gcd(long long v, long long b) {
  if (v == 0) return b;
  if (b == 0) return v;
  if (v > b) return gcd(b, v);
  if (v == b) return b;
  if (b % v == 0) return v;
  return gcd(v, b % v);
}
long long mod;
long long extgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long d = extgcd(b, a % b, y, x);
  y -= a / b * x;
  return d;
}
pair<long long, long long> operator+(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first + r.first, l.second + r.second};
}
pair<long long, long long> operator-(const pair<long long, long long> &l,
                                     const pair<long long, long long> &r) {
  return {l.first - r.first, l.second - r.second};
}
long long pr[1000010];
long long inv[1000010];
long long beki(long long wa, long long rr, long long warukazu) {
  if (rr == 0) return 1 % warukazu;
  if (rr == 1) return wa % warukazu;
  wa %= warukazu;
  if (rr % 2 == 1)
    return ((long long)beki(wa, rr - 1, warukazu) * (long long)wa) % warukazu;
  long long zx = beki(wa, rr / 2, warukazu);
  return (zx * zx) % warukazu;
}
long long comb(long long nn, long long rr) {
  if (rr < 0 || rr > nn || nn < 0) return 0;
  long long r = pr[nn] * inv[rr];
  r %= mod;
  r *= inv[nn - rr];
  r %= mod;
  return r;
}
void gya(long long ert) {
  pr[0] = 1;
  for (long long i = 1; i <= ert; i++) {
    pr[i] = (pr[i - 1] * i) % mod;
  }
  inv[ert] = beki(pr[ert], mod - 2, mod);
  for (long long i = ert - 1; i >= 0; i--) {
    inv[i] = inv[i + 1] * (i + 1) % mod;
  }
}
long long a[100020][3];
vector<long long> ve;
vector<long long> G[100020];
void dfs(long long r, long long p) {
  ve.push_back(r);
  for (auto v : G[r])
    if (v != p) dfs(v, r);
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  map<long long, long long> ma;
  for (long long i = 0; i < n; i++) {
    long long y;
    cin >> y;
    ma[y]++;
  }
  deque<pair<long long, long long> > de;
  for (auto it = ma.begin(); it != ma.end(); it++) {
    pair<long long, long long> z = *it;
    de.push_back(z);
  }
  while (de.size() > 1) {
    n = de.size();
    pair<long long, long long> a = de.front();
    pair<long long, long long> b = de.back();
    if (k == 0) break;
    if (a.second < b.second) {
      long long sa = de[1].first - de[0].first;
      if (k < a.second * sa) {
        de[0].first += k / a.second;
        break;
      } else {
        k -= a.second * sa;
        de.pop_front();
        de[0].second += a.second;
      }
    } else {
      long long sa = de[n - 1].first - de[n - 2].first;
      if (k < b.second * sa) {
        de[n - 1].first -= k / b.second;
        break;
      } else {
        k -= b.second * sa;
        de.pop_back();
        de.back().second += b.second;
      }
    }
  }
  cout << -de[0].first + de.back().first << endl;
  return 0;
}

```