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>
#pragma GCC optimize 2
using namespace std;
const int maxn = 1E3 + 5;
int n, k, a[maxn], maxx, where[maxn];
long long int f[maxn][maxn], sum[maxn][maxn];
long long int get(int x) {
  for (int i = 0; i <= k; ++i)
    for (int j = 0; j <= n; ++j) f[i][j] = 0;
  for (int i = 1; i <= n; ++i) {
    where[i] = where[i - 1];
    while (a[i] - a[where[i] + 1] >= x) ++where[i];
  }
  f[0][0] = 1;
  for (int i = 0; i < k; ++i) {
    sum[i][0] = f[i][0];
    for (int j = 1; j <= n; ++j)
      sum[i][j] = (sum[i][j - 1] + f[i][j]) % 998244353;
    for (int j = 1; j <= n; ++j) f[i + 1][j] = sum[i][where[j]];
  }
  long long int sum = 0;
  for (int i = 1; i <= n; ++i) sum = (sum + f[k][i]) % 998244353;
  return sum;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    maxx = max(maxx, a[i]);
  }
  sort(a + 1, a + n + 1);
  long long int ans = 0;
  for (int i = 1; i <= 100000 / (k - 1) + 10; ++i)
    ans = (ans + get(i)) % 998244353;
  cout << ans << endl;
  return 0;
}
