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 long long mod = 998244353;
const int maxn = 1010;
int dp[maxn][maxn];
int sum[maxn][maxn];
int a[maxn];
int ans = 0;
int n, k;
void solve(int x) {
  for (int i = 0; i <= n; i++)
    for (int j = 1; j <= k; j++) {
      dp[i][j] = 0;
      sum[i][j] = 0;
    }
  for (int i = 1; i <= n; i++) dp[i][1] = 1;
  for (int j = 1; j <= k; j++) {
    int l = 1;
    for (int i = 1; i <= n; i++) {
      while (a[i] - a[l] >= x) l++;
      if (j == 1)
        dp[i][j] = (dp[i][j] + sum[l - 1][j - 1]) % mod;
      else
        dp[i][j] = (sum[l - 1][j - 1]) % mod;
      sum[i][j] = (dp[i][j] + sum[i - 1][j]) % mod;
    }
  }
  for (int i = 1; i <= n; i++) ans = (ans + dp[i][k]) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + 1 + n);
  for (int i = 1; i <= a[n] / (k - 1); i++) {
    solve(i);
  }
  printf("%d\n", ans);
}
