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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long dp[1005][1005], a[1005], b[1005], cum[1005];
long long b1[1005];
long long cal(long long x, long long n, long long k) {
  long long i, j = n - 1;
  for (i = n; i >= 1; i--) {
    dp[i][1] = i;
    b[i] = 0;
    while (j >= 1) {
      if (a[i] - a[j] >= x) {
        b[i] = j;
        break;
      } else
        j--;
    }
  }
  long long s = 0;
  for (i = 2; i <= k; i++) {
    for (j = 1; j <= n; j++) {
      long long p = b[j];
      dp[j][i] = dp[p][i - 1] + dp[j - 1][i];
      dp[j][i] %= 998244353;
    }
  }
  return dp[n][k];
}
int main() {
  long long n, k, i, j;
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  while (cin >> n >> k) {
    for (i = 1; i <= n; i++) cin >> a[i];
    sort(a + 1, a + n + 1);
    long long m = a[n] / (k - 1) + 1;
    long long ans = 0;
    long long s = 0;
    long long pre = 0;
    for (i = 100000; i >= 0; i--) {
      if (a[1] + i * (k - 1) > a[n]) continue;
      long long p = cal(i, n, k);
      ans += (p - pre + 998244353) * i;
      ans %= 998244353;
      pre = p;
    }
    cout << ans << endl;
  }
  return 0;
}
