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 = 1e3 + 10;
int n, k, ans, a[maxn], pre[maxn];
int dp[maxn][maxn], p[maxn];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  sort(a + 1, a + n + 1);
  dp[0][0] = 1;
  for (int l = 1; l * (k - 1) < 1e5 + 10; l++) {
    for (int i = 1; i <= n; i++) {
      p[i] = p[i - 1];
      while (a[i] - a[p[i] + 1] >= l) p[i]++;
    }
    for (int i = 0; i + 1 <= k; i++) {
      pre[0] = dp[i][0];
      for (int j = 1; j <= n; j++) pre[j] = (pre[j - 1] + dp[i][j]) % mod;
      for (int j = 1; j <= n; j++) dp[i + 1][j] = pre[p[j]];
    }
    for (int i = 1; i <= n; i++) (ans += dp[k][i]) %= mod;
  }
  printf("%d\n", ans);
}
