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, N = 1005, M = N * 100, inf = 0x3f3f3f3f;
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int sub(int a, int b) { return a >= b ? a - b : a - b + mod; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }
int n, k, ans, a[N], f[N][N], sum[N][N], res[M];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  a[0] = -inf;
  int Max = a[n];
  for (int val = 1; val * (k - 1) <= Max; val++) {
    sum[0][0] = f[0][0] = 1;
    int pos = 0;
    for (int i = 1; i <= n; i++) {
      while (a[pos] <= a[i] - val) pos++;
      for (int j = 0; j <= k; j++) {
        if (j) f[i][j] = sum[pos - 1][j - 1];
        sum[i][j] = add(sum[i - 1][j], f[i][j]);
      }
      res[val] = add(f[i][k], res[val]);
    }
    ans = add(ans, res[val]);
  }
  cout << ans << endl;
  return 0;
}
