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;
long long n, K;
long long dp[1005][1005][2], sum[2][1005];
long long a[1005], ub[100005], lb[100005];
bool used[100005];
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> K;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) used[a[i]] = true;
  for (int i = 0; i < 100005; i++) {
    if (!used[i]) {
      ub[i] = lb[i] = ub[i - 1];
      if (i == 0) ub[i] = lb[i] = 0;
      continue;
    }
    ub[i] = upper_bound(a + 1, a + n + 1, i) - a - 1;
    lb[i] = lower_bound(a + 1, a + n + 1, i) - a - 1;
  }
  long long m = 100000 / (K - 1), ans = 0;
  for (int d = 1; d <= m; d++) {
    for (int i = 1; i <= K; i++) {
      for (int j = 0; j <= n; j++) {
        for (int k = 0; k < 2; k++) {
          dp[i][j][k] = 0;
        }
      }
    }
    for (int j = 1; j <= n; j++) dp[1][j][0] = 1;
    for (int i = 2; i <= K; i++) {
      for (int j = 1; j <= n; j++) {
        sum[0][j] = (sum[0][j - 1] + dp[i - 1][j][0]) % 998244353;
        sum[1][j] = (sum[1][j - 1] + dp[i - 1][j][1]) % 998244353;
      }
      for (int j = 1; j <= n; j++) {
        long long just = a[j] - d;
        if (just >= 0)
          dp[i][j][0] += sum[0][lb[just]], dp[i][j][0] %= 998244353;
        if (just >= 0) {
          dp[i][j][1] += sum[1][ub[just]], dp[i][j][1] %= 998244353;
          dp[i][j][1] += sum[0][ub[just]] - sum[0][lb[just]] + 998244353,
              dp[i][j][1] %= 998244353;
        }
      }
    }
    for (int j = 1; j <= n; j++)
      ans += d * dp[K][j][1] % 998244353, ans %= 998244353;
  }
  cout << ans << endl;
  return 0;
}
