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>
using namespace std;
void debug_out() { cerr << endl; }
template <class T>
ostream& prnt(ostream& out, T v) {
  out << v.size() << '\n';
  for (auto e : v) out << e << ' ';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> v) {
  return prnt(out, v);
}
template <class T>
ostream& operator<<(ostream& out, set<T> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ' ' << p.second << ')';
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
const int N = 1001;
const int MOD = 998244353;
int n, dp[2][N], a[N], k, f[100100], r[N];
int solve(int x) {
  for (int i = 1; i <= n; i++) {
    int j = upper_bound(a + 1, a + n + 1, a[i] - x) - a;
    j--;
    r[i] = j;
  }
  for (int j = 1; j <= k; j++) {
    for (int i = 1; i <= n; i++)
      if (j == 1)
        dp[1][i] = 1;
      else
        dp[j % 2][i] = dp[1 - j % 2][r[i]];
    for (int i = 2; i <= n; i++)
      dp[j % 2][i] += dp[j % 2][i - 1], dp[j % 2][i] %= MOD;
  }
  return dp[k % 2][n];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  int mx = a[n] / (k - 1) + 1;
  for (int x = 0; x <= mx; x++) f[x] = solve(x);
  long long ans = 0;
  for (int x = 0; x < mx; x++) {
    f[x] -= f[x + 1];
    if (f[x] < 0) f[x] += MOD;
    ans += (1LL * x * f[x]) % MOD;
    ans %= MOD;
  }
  cout << ans << '\n';
}
