Let's call beauty of an array b_1, b_2, …, b_n (n > 1) — min_{1 ≤ i < j ≤ n} |b_i - b_j|.

You're given an array a_1, a_2, … a_n and a number k. Calculate the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.

A sequence a is a subsequence of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements.

Input

The first line contains integers n, k (2 ≤ k ≤ n ≤ 1000).

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^5).

Output

Output one integer — the sum of beauty over all subsequences of the array of length exactly k. As this number can be very large, output it modulo 998244353.

Examples

Input


4 3
1 7 3 5


Output


8

Input


5 5
1 10 100 1000 10000


Output


9

Note

In the first example, there are 4 subsequences of length 3 — [1, 7, 3], [1, 3, 5], [7, 3, 5], [1, 7, 5], each of which has beauty 2, so answer is 8.

In the second example, there is only one subsequence of length 5 — the whole array, which has the beauty equal to |10-1| = 9.
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#pragma GCC target("avx2")
using namespace std;
template <class T, class U>
inline void checkmin(T &x, U y) {
  if (y < x) x = y;
}
template <class T, class U>
inline void checkmax(T &x, U y) {
  if (y > x) x = y;
}
template <class T, class U>
inline bool ifmax(T &x, U y) {
  if (y > x) return x = y, true;
  return false;
}
template <class T, class U>
inline bool ifmin(T &x, U y) {
  if (y < x) return x = y, true;
  return false;
}
template <class T>
inline void sort(T &a) {
  sort(a.begin(), a.end());
}
template <class T>
inline void rsort(T &a) {
  sort(a.rbegin(), a.rend());
}
template <class T>
inline void reverse(T &a) {
  reverse(a.begin(), a.end());
}
template <class T, class U>
inline istream &operator>>(istream &str, pair<T, U> &p) {
  return str >> p.first >> p.second;
}
template <class T>
inline istream &operator>>(istream &str, vector<T> &a) {
  for (auto &i : a) str >> i;
  return str;
}
struct zet {
  int val;
  zet(long long x = 0) {
    val = x % 998244353;
    if (val < 0) val += 998244353;
  }
  zet(long long a, long long b) {
    *this += a;
    *this /= b;
  }
  zet &operator+=(zet const &b) {
    val += b.val;
    if (val >= 998244353) val -= 998244353;
    return *this;
  }
  zet &operator-=(zet const &b) {
    val -= b.val;
    if (val < 0) val += 998244353;
    return *this;
  }
  zet &operator*=(zet const &b) {
    val = (val * (long long)b.val) % 998244353;
    return *this;
  }
  friend zet mypow(zet a, long long n) {
    zet res = 1;
    while (n) {
      if (n & 1) res *= a;
      a *= a;
      n >>= 1;
    }
    return res;
  }
  friend zet inv(zet a) { return mypow(a, 998244353 - 2); }
  zet &operator/=(zet const &b) { return *this *= inv(b); }
  friend zet operator+(zet a, const zet &b) { return a += b; }
  friend zet operator-(zet a, const zet &b) { return a -= b; }
  friend zet operator-(zet a) { return 0 - a; }
  friend zet operator*(zet a, const zet &b) { return a *= b; }
  friend zet operator/(zet a, const zet &b) { return a /= b; }
  friend istream &operator>>(istream &stream, zet &a) {
    return stream >> a.val;
  }
  friend ostream &operator<<(ostream &stream, const zet &a) {
    return stream << a.val;
  }
  friend bool operator==(zet const &a, zet const &b) { return a.val == b.val; }
  friend bool operator!=(zet const &a, zet const &b) { return a.val != b.val; }
  friend bool operator<(zet const &a, zet const &b) { return a.val < b.val; }
};
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  srand(94385);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  cin >> a;
  sort(a);
  zet ans = 0;
  for (int d = 1;; ++d) {
    vector<vector<zet>> dp(k, vector<zet>(n));
    dp[0] = vector<zet>(n, 1);
    for (int j = 1; j < k; ++j) {
      int ptr = 0;
      for (int i = 0; i < n - 1; ++i) {
        checkmax(ptr, i + 1);
        while (ptr < n && a[ptr] - a[i] < d) ++ptr;
        if (ptr == n) break;
        dp[j][ptr] += dp[j - 1][i];
      }
      for (int i = 1; i < n; ++i) dp[j][i] += dp[j][i - 1];
    }
    zet add = 0;
    for (int i = 0; i < n; ++i) add += dp[k - 1][i];
    if (add == 0) break;
    ans += add;
  }
  cout << ans;
  return 0;
}
