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 maxn = 1010, P = 998244353;
int n, K, ans, a[maxn], cur[maxn], f[maxn], g[maxn];
void red(int &x) { x += x >> 31 & P; }
int main() {
  scanf("%d %d", &n, &K);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]), cur[i] = i - 1;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i * (K - 1) <= 100000; i++) {
    memset(g, 0, sizeof(g));
    for (int k = 1; k <= K; k++) {
      for (int j = 1; j <= n; j++) {
        while (cur[j] && a[j] - a[cur[j]] < i) cur[j]--;
        f[j] = k == 1 ? 1 : g[cur[j]];
      }
      for (int j = 1; j <= n; j++) {
        g[j] = g[j - 1], red(g[j] += f[j] - P);
      }
    }
    red(ans += g[n] - P);
  }
  printf("%d\n", ans);
  return 0;
}
