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;
inline int read() {
  int x = 0, ch = getchar();
  bool fg = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') fg = 0;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return fg ? x : -x;
}
const int MAXN = 1005, mod = 998244353;
int a[MAXN], dp[MAXN][MAXN];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  int mx = a[n] / (k - 1) + 1;
  int ans = 0;
  for (int i = 1; i <= k; i++) dp[0][i] = -1;
  int last = 0;
  for (int x = mx; x > 0; x--) {
    int cur = 0;
    for (int i = 1; i <= n; i++) {
      while (cur + 1 < i && a[i] - a[cur + 1] >= x) cur++;
      for (int j = 1; j <= k; j++) {
        dp[i][j] = (dp[cur][j - 1] + 1 + dp[i - 1][j]) % mod;
      }
    }
    if (~dp[n][k]) {
      ans = (ans + (long long)(dp[n][k] + 1 - last) * x) % mod;
      last = dp[n][k] + 1;
    }
  }
  cout << (ans + mod) % mod << endl;
  return 0;
}
