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;
const int maxn = 5001;
int n, k;
int a[maxn], dp[maxn][maxn], sum[maxn][maxn], MAX;
long long ans;
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), MAX = max(MAX, a[i]);
  sort(a + 1, a + n + 1);
  for (int x = 1; x <= MAX; ++x) {
    if (x > 1)
      for (int i = 0; i <= n; ++i)
        for (int j = 0; j <= min(n, MAX / (x - 1) + 1); ++j) dp[i][j] = 0;
    int top = 1;
    sum[0][0] = dp[0][0] = 1;
    for (int i = 1; i <= n; ++i) {
      while (top <= i && a[i] - a[top] >= x) {
        for (int j = 0; j <= min(n, MAX / x + 1); ++j) {
          sum[top][j] = (sum[top - 1][j] + dp[top][j]);
          if (sum[top][j] >= mod) sum[top][j] -= mod;
        }
        ++top;
      }
      int qwq = top - 1;
      for (int j = 1; j <= min(n, MAX / x + 1); ++j) {
        dp[i][j] = (dp[i][j] + sum[qwq][j - 1]);
        if (dp[i][j] >= mod) dp[i][j] -= mod;
      }
      ans = (ans + dp[i][k]);
      if (ans >= mod) ans -= mod;
    }
  }
  cout << ans << "\n";
  return 0;
}
