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 INF = 0x3f3f3f3f;
const long long mo = 998244353;
long long n, k, ans;
long long a[1005], res[100005];
long long dp[1005][1005], g[1005][1005];
int main() {
  scanf("%lld %lld", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + n + 1);
  a[0] = -INF;
  for (int i = 1; i * (k - 1) <= a[n]; i++) {
    dp[0][0] = g[0][0] = 1;
    int now = 0;
    for (int j = 1; j <= n; j++) {
      while (a[now] <= a[j] - i) now++;
      for (int l = 0; l <= k; l++) {
        if (l) dp[j][l] = g[now - 1][l - 1];
        g[j][l] = (g[j - 1][l] + dp[j][l]) % mo;
      }
      res[i] += dp[j][k];
    }
    ans = (ans + res[i]) % mo;
  }
  printf("%lld", ans);
  return 0;
}
