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 MAXN = 1050;
const long long INF = 0x3f3f3f3f;
const long long LINF = 0x1f3f3f3f3f3f3f3f;
const long long MOD = (long long)998244353;
const long long OVER_FLOW = 0xffffffff;
long long n;
long long a[MAXN], f[MAXN][MAXN], res[100050], s[MAXN][MAXN];
int main() {
  ios::sync_with_stdio(false);
  long long k;
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  a[0] = -INF;
  long long ans = 0;
  for (long long v = 1; v * (k - 1) <= a[n]; v++) {
    f[0][0] = 1;
    long long now = 0;
    s[0][0] = 1;
    for (long long i = 1; i <= n; i++) {
      while (a[now] <= a[i] - v) now++;
      for (long long j = 0; j <= k; j++) {
        if (j) f[i][j] = s[now - 1][j - 1];
        s[i][j] = (s[i - 1][j] + f[i][j]) % MOD;
      }
      res[v] += f[i][k];
    }
    ans += res[v] % MOD;
  }
  cout << ans % MOD << endl;
  return 0;
}
