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(3)
using namespace std;
long long mod = 998244353;
const long long N = 2e5 + 10;
long long n, a[N];
long long dp[1020][1020];
long long pre[1020][1020];
long long K;
long long get(long long x) {
  long long k = 1;
  for (long long i = 1; i <= n; i++) {
    for (; k <= i; k++)
      if (a[i] < a[k] + x) break;
    for (long long j = 2; j <= K; j++) {
      dp[i][j] = pre[k - 1][j - 1];
      pre[i][j] = pre[i - 1][j] + dp[i][j];
      if (pre[i][j] >= mod) pre[i][j] -= mod;
    }
  }
  return pre[n][K];
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> K;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  for (long long i = 1; i <= n; i++) dp[i][1] = 1;
  for (long long i = 1; i <= n; i++) pre[i][1] = i;
  long long ans = 0;
  long long mx = 1e5 / (K - 1);
  for (long long i = 1; i <= mx; i++) {
    ans += get(i);
    ans %= mod;
  }
  cout << ans << endl;
}
