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 long long maxx = 100000 + 7;
const long long maxn = 1000 + 7;
const long long mod = 998244353;
const long long maxs = 5 * (1E5) + 7;
const long long maxm = 30000 + 7;
const long long maxk = 800 + 7;
const long long maxw = 100 + 7;
const long long maxh = 100 + 7;
const double inf = 1e18 + 7;
ostream& operator<<(ostream& os, pair<long long, long long>& rhs) {
  os << rhs.first << " " << rhs.second << endl;
  return os;
}
long long seq[maxn];
long long dp[maxn][maxn];
long long n, k;
long long ans[maxx];
long long lim;
void init() {
  scanf("%lld%lld", &n, &k);
  for (long long i = (1); i <= (n); i++) scanf("%lld", &seq[i]);
  sort(seq + 1, seq + n + 1);
  lim = (seq[n] / (k - 1) + 1);
}
void solve() {
  long long xs = 0;
  for (long long x = (0); x <= (lim); x++) {
    long long p = 1;
    for (long long i = (2); i <= (n); i++) {
      while (p < i && seq[i] - seq[p] >= x) p++;
      dp[i][2] = dp[i - 1][2] + p - 1;
      for (long long j = (3); j <= (min(k, i)); j++)
        dp[i][j] = (dp[i - 1][j] + dp[p - 1][j - 1]) % mod;
    }
    ans[x] = dp[n][k];
    if (x) {
      long long tmp = (ans[x - 1] - ans[x] + mod) % mod;
      long long tt = tmp * (x - 1) % mod;
      xs += tt;
      xs %= mod;
    }
  }
  printf("%lld\n", xs);
}
int main(void) {
  init();
  solve();
  return 0;
}
