#include <bits/stdc++.h>
using namespace std;
const long long maxn = 500005;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const long long MOD = 1e9 + 7;
const double eps = 1e-10;
const double pi = 3.1415926535;
long long qpow(long long a, long long b) {
  long long tmp = a % MOD, ans = 1;
  while (b) {
    if (b & 1) {
      ans *= tmp, ans %= MOD;
    }
    tmp *= tmp, tmp %= MOD, b >>= 1;
  }
  return ans;
}
long long lowbit(long long x) { return x & -x; }
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
long long mmax(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long mmin(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
void mod(long long &a) {
  a += MOD;
  a %= MOD;
}
bool chk(long long now);
long long half(long long l, long long r) {
  while (l <= r) {
    long long m = (l + r) / 2;
    if (chk(m))
      r = m - 1;
    else
      l = m + 1;
  }
  return l;
}
long long ll(long long p) { return p << 1; }
long long rr(long long p) { return p << 1 | 1; }
long long mm(long long l, long long r) { return (l + r) / 2; }
long long lg(long long x) {
  if (x == 0) return 1;
  return (long long)log2(x) + 1;
}
bool smleql(double a, double b) {
  if (a < b || fabs(a - b) <= eps) return true;
  return false;
}
bool bigeql(double a, double b) {
  if (a > b || fabs(a - b) <= eps) return true;
  return false;
}
bool eql(double a, double b) {
  if (fabs(a - b) < eps) return 1;
  return 0;
}
double len(double a, double b, double c, double d) {
  return sqrt((a - c) * (a - c) + (b - d) * (b - d));
}
bool isp(long long x) {
  if (x == 1) return false;
  if (x == 2) return true;
  for (long long i = 2; i * i <= x; ++i)
    if (x % i == 0) return false;
  return true;
}
inline long long read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < 48 || ch > 57) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57) {
    s = (s << 1) + (s << 3) + ch - 48;
    ch = getchar();
  }
  return s * w;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (!(a & 1) && !(b & 1))
    return gcd(a >> 1, b >> 1) << 1;
  else if (!(b & 1))
    return gcd(a, b >> 1);
  else if (!(a & 1))
    return gcd(a >> 1, b);
  else
    return gcd(abs(a - b), min(a, b));
}
long long lcm(long long x, long long y) { return x * y / gcd(x, y); }
long long n, k, a[maxn], dp[maxn];
bool chk(long long now) {
  memset(dp, 0, sizeof(dp));
  dp[0] = 1;
  long long last = 1;
  for (long long i = 1; i <= n; ++i) {
    while (a[i] - a[last] > now) last++;
    while (i - last + 1 >= k) {
      if (dp[last - 1]) {
        dp[i] = 1;
        break;
      }
      last++;
    }
  }
  return dp[n];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  n = read(), k = read();
  for (long long i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + 1 + n);
  cout << half(0, 1e18);
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

Evlampiy has found one more cool application to process photos. However the application has certain limitations.

Each photo i has a contrast vi. In order for the processing to be truly of high quality, the application must receive at least k photos with contrasts which differ as little as possible.

Evlampiy already knows the contrast vi for each of his n photos. Now he wants to split the photos into groups, so that each group contains at least k photos. As a result, each photo must belong to exactly one group.

He considers a processing time of the j-th group to be the difference between the maximum and minimum values of vi in the group. Because of multithreading the processing time of a division into groups is the maximum processing time among all groups.

Split n photos into groups in a such way that the processing time of the division is the minimum possible, i.e. that the the maximum processing time over all groups as least as possible.

Input

The first line contains two integers n and k (1 ≤ k ≤ n ≤ 3·105) — number of photos and minimum size of a group.

The second line contains n integers v1, v2, ..., vn (1 ≤ vi ≤ 109), where vi is the contrast of the i-th photo.

Output

Print the minimal processing time of the division into groups.

Examples

Input

5 2
50 110 130 40 120


Output

20


Input

4 1
2 3 4 1


Output

0

Note

In the first example the photos should be split into 2 groups: [40, 50] and [110, 120, 130]. The processing time of the first group is 10, and the processing time of the second group is 20. Maximum among 10 and 20 is 20. It is impossible to split the photos into groups in a such way that the processing time of division is less than 20.

In the second example the photos should be split into four groups, each containing one photo. So the minimal possible processing time of a division is 0.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long maxn = 500005;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const long long MOD = 1e9 + 7;
const double eps = 1e-10;
const double pi = 3.1415926535;
long long qpow(long long a, long long b) {
  long long tmp = a % MOD, ans = 1;
  while (b) {
    if (b & 1) {
      ans *= tmp, ans %= MOD;
    }
    tmp *= tmp, tmp %= MOD, b >>= 1;
  }
  return ans;
}
long long lowbit(long long x) { return x & -x; }
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
long long mmax(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long mmin(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
void mod(long long &a) {
  a += MOD;
  a %= MOD;
}
bool chk(long long now);
long long half(long long l, long long r) {
  while (l <= r) {
    long long m = (l + r) / 2;
    if (chk(m))
      r = m - 1;
    else
      l = m + 1;
  }
  return l;
}
long long ll(long long p) { return p << 1; }
long long rr(long long p) { return p << 1 | 1; }
long long mm(long long l, long long r) { return (l + r) / 2; }
long long lg(long long x) {
  if (x == 0) return 1;
  return (long long)log2(x) + 1;
}
bool smleql(double a, double b) {
  if (a < b || fabs(a - b) <= eps) return true;
  return false;
}
bool bigeql(double a, double b) {
  if (a > b || fabs(a - b) <= eps) return true;
  return false;
}
bool eql(double a, double b) {
  if (fabs(a - b) < eps) return 1;
  return 0;
}
double len(double a, double b, double c, double d) {
  return sqrt((a - c) * (a - c) + (b - d) * (b - d));
}
bool isp(long long x) {
  if (x == 1) return false;
  if (x == 2) return true;
  for (long long i = 2; i * i <= x; ++i)
    if (x % i == 0) return false;
  return true;
}
inline long long read() {
  char ch = getchar();
  long long s = 0, w = 1;
  while (ch < 48 || ch > 57) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= 48 && ch <= 57) {
    s = (s << 1) + (s << 3) + ch - 48;
    ch = getchar();
  }
  return s * w;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (!(a & 1) && !(b & 1))
    return gcd(a >> 1, b >> 1) << 1;
  else if (!(b & 1))
    return gcd(a, b >> 1);
  else if (!(a & 1))
    return gcd(a >> 1, b);
  else
    return gcd(abs(a - b), min(a, b));
}
long long lcm(long long x, long long y) { return x * y / gcd(x, y); }
long long n, k, a[maxn], dp[maxn];
bool chk(long long now) {
  memset(dp, 0, sizeof(dp));
  dp[0] = 1;
  long long last = 1;
  for (long long i = 1; i <= n; ++i) {
    while (a[i] - a[last] > now) last++;
    while (i - last + 1 >= k) {
      if (dp[last - 1]) {
        dp[i] = 1;
        break;
      }
      last++;
    }
  }
  return dp[n];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  n = read(), k = read();
  for (long long i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + 1 + n);
  cout << half(0, 1e18);
  return 0;
}

```