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 q = 0, w = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (isdigit(c)) q = q * 10 + c - '0', c = getchar();
  return w * q;
}
const int N = 1010;
const int mod = 998244353;
int n, k, ans, a[N], dp[N][N], sum[N][N];
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + 1 + n);
  for (int gap = 1; gap <= (a[n] - a[1]) / (k - 1); ++gap) {
    for (int i = 0; i <= n; ++i)
      for (int j = 0; j <= k; ++j) dp[i][j] = sum[i][j] = 0;
    for (int i = 1; i <= n; ++i) dp[i][1] = 1, sum[i][1] = i;
    for (int j = 2; j <= k; ++j) {
      int cur = 0;
      for (int i = 1; i <= n; ++i) {
        while (cur < i && a[i] - a[cur + 1] >= gap) cur++;
        dp[i][j] = sum[cur][j - 1];
        sum[i][j] = (sum[i - 1][j] + dp[i][j]) % mod;
      }
    }
    ans = (ans + sum[n][k]) % mod;
  }
  cout << ans << endl;
  return 0;
}
