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 N = 3e5 + 10;
const int mod = 998244353;
int up(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
int a[1010], dp[1010][1010], pos[100010], n, k;
int cal(int x) {
  for (int i = 0; i <= k; ++i)
    for (int j = 0; j <= n; ++j) dp[i][j] = 0;
  for (int i = 1; i <= n; ++i) dp[1][i] = 1;
  for (int i = 2; i <= k + 1; ++i) {
    for (int j = 1; j <= n; ++j)
      dp[i - 1][j] = up(dp[i - 1][j - 1], dp[i - 1][j]);
    for (int j = 1; j <= n; ++j)
      dp[i][j] = dp[i - 1][a[j] >= x ? pos[a[j] - x] : 0];
  }
  return dp[k][n];
}
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 = n; i >= 1; --i)
    pos[a[i]] = (a[i] == a[i + 1] ? pos[a[i + 1]] : i);
  for (int i = 1; i <= 100000; ++i) pos[i] = max(pos[i], pos[i - 1]);
  int ans = 0;
  for (int i = 1; i * (k - 1) <= 100000; ++i) ans = up(ans, cal(i));
  printf("%d\n", ans);
  return 0;
}
