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;
const int Maxn = 2e5 + 10;
long long Inf = 1e18;
const int Log = 20;
const long long Sqrt = 1000000;
const long long Mod = 998244353LL;
long long mul(long long a, long long b) {
  a %= Mod;
  b %= Mod;
  return (a * b) % Mod;
}
long long bin_pow(long long b, long long p) {
  long long res = 1;
  for (long long j = 1, pw = b; j <= p; j <<= 1, pw = mul(pw, pw))
    if (p & j) res = mul(res, pw);
  return res;
}
long long a[Maxn];
long long X = 110000;
long long cnt[Maxn];
long long dp[1010][1010];
long long ps[1010][1010];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  a[0] = -10000000;
  sort(a + 1, a + n + 1);
  long long mx = X / max(1LL, (k - 1));
  for (int i = 1; i <= mx; i++) {
    for (int ii = 1; ii <= n; ii++)
      for (int j = 1; j <= k; j++) dp[ii][j] = 0, ps[ii][j] = 0;
    long long p1 = 0;
    dp[0][0] = 1;
    ps[0][0] = 1;
    for (int j = 1; j <= n; j++) ps[j][0] = 1;
    for (int j = 1; j <= n; j++) {
      while (a[p1] + i <= a[j]) {
        p1++;
      }
      p1--;
      for (int K = 1; K <= k; K++) {
        dp[j][K] = ps[p1][K - 1];
        ps[j][K] = (ps[j - 1][K] + dp[j][K]) % Mod;
      }
      cnt[i] += dp[j][k];
      cnt[i] %= Mod;
    }
  }
  long long ans = 0;
  for (int i = 1; i <= mx; i++) {
    ans += mul(i, cnt[i] - cnt[i + 1]);
    ans %= Mod;
  }
  ans += Mod;
  cout << ans % Mod;
  return 0;
}
