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 MOD = 998244353;
inline int add(const int& x, const int& y) {
  return x + y >= MOD ? x + y - MOD : x + y;
}
inline int dec(const int& x, const int& y) {
  return x < y ? x + MOD - y : x - y;
}
int a[1005];
int dp[1005][1005], ans[100005];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  int mx = (a[n] - a[1]) / (k - 1);
  for (int d = 1; d <= mx; d++) {
    for (int i = 1; i <= k; i++)
      for (int j = 1; j <= n; j++) dp[i][j] = 0;
    for (int i = 1; i <= n; i++) dp[1][i] = 1;
    for (int i = 2; i <= k; i++) {
      int pos = 0, sum = 0;
      for (int j = 1; j <= n; j++) {
        while (pos < n && a[j] - a[pos + 1] >= d)
          sum = add(sum, dp[i - 1][++pos]);
        dp[i][j] = sum;
      }
    }
    for (int i = 1; i <= n; i++) ans[d] = add(ans[d], dp[k][i]);
  }
  int res = 0;
  for (int i = 1; i <= mx; i++) res = add(res, ans[i]);
  printf("%d\n", res);
  return 0;
}
