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 = 1009;
const int mod = 998244353;
int a[N], dp[N][N], sum[N];
int n, k;
int ans;
void add(int x, int y) {
  for (; x <= n; x += (x & -x)) {
    sum[x] = (sum[x] + y);
    if (sum[x] >= mod) sum[x] -= mod;
  }
}
int query(int x) {
  int ans = 0;
  while (x) {
    ans = (ans + sum[x]);
    if (ans >= mod) ans -= mod;
    x -= (x & -x);
  }
  return ans;
}
void solve(int x) {
  for (int j = 2; j <= k; j++) {
    for (int i = 1; i <= n; i++) {
      int l = upper_bound(a + 1, a + n + 1, a[i] - x) - a - 1;
      dp[i][j] = sum[l];
    }
    for (int i = 1; i <= n; i++) {
      sum[i] = dp[i][j] + sum[i - 1];
      if (sum[i] > mod) sum[i] -= mod;
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  int limit = a[n] / (k - 1);
  for (int x = 1; x <= limit; x++) {
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= k; j++) dp[i][j] = 0;
    for (int i = 0; i <= n; i++) sum[i] = 0;
    for (int i = 1; i <= n; i++) dp[i][1] = 1, sum[i] = i;
    solve(x);
    for (int i = k; i <= n; i++) {
      ans = (ans + dp[i][k]);
      if (ans >= mod) ans -= mod;
    }
  }
  if (ans < 0) ans += mod;
  cout << ans << endl;
  return 0;
}
