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>
const int N = 1e3 + 7;
const int MOD = 998244353;
int a[N], dp[N][N], sum[N][N];
int n, K;
int M(int x) {
  if (x >= MOD) x -= MOD;
  if (x < 0) x += MOD;
  return x;
}
int cal(int limit) {
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= K; j++) dp[i][j] = sum[i][j] = 0;
  for (int i = 1; i <= n; i++) dp[i][1] = 1, sum[i][1] = i;
  for (int k = 2; k <= K; k++) {
    int pos = 0;
    for (int i = 1; i <= n; i++) {
      while (pos < i && a[i] - a[pos] >= limit) pos++;
      if (pos && a[i] - a[pos] < limit) pos--;
      dp[i][k] = sum[pos][k - 1];
      sum[i][k] = M(sum[i - 1][k] + dp[i][k]);
    }
  }
  return sum[n][K];
}
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  std::sort(a + 1, a + n + 1);
  int ans = 0;
  for (int i = 1; i <= (a[n] - a[1]) / (K - 1); i++) ans = M(ans + cal(i));
  printf("%d\n", ans);
  return 0;
}
