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 int N = 1005, P = 998244353, M = 100005;
int n, kk, a[N];
int dp[N][N], sum[N][N], l[N];
inline void add(long long &a, long long b) { a = (a + b % P) % P; }
inline void add(int &a, long long b) { a = (a + b % P) % P; }
int solve(int mx) {
  for (int i = 1; i <= kk; i++)
    for (int j = 1; j <= n; j++) sum[i][j] = dp[i][j] = 0;
  for (int i = 1; i <= n; i++) {
    dp[1][i] = 1;
    l[i] = l[i - 1];
    while (a[i] - a[l[i] + 1] >= mx) ++l[i];
  }
  for (int i = 1; i <= kk; i++)
    for (int j = 1; j <= n; j++) {
      add(dp[i][j], sum[i - 1][l[j]]);
      sum[i][j] = (sum[i][j - 1] + dp[i][j]) % P;
    }
  int ans = 0;
  for (int i = 1; i <= n; i++) add(ans, dp[kk][i]);
  return ans;
}
int main() {
  scanf("%d%d", &n, &kk);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  long long ans = 0;
  for (int i = 1; i <= a[n] / (kk - 1); i++) add(ans, solve(i));
  printf("%I64d\n", ans);
  return 0;
}
