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 = 1010;
int n, K, mxa;
int a[maxn];
inline int MOD(int x) { return x < mod ? x : x - mod; }
inline int Mod(int x) { return x < 0 ? x + mod : x; }
void input() {
  int i;
  scanf("%d%d", &n, &K);
  for (i = 1; i <= n; i++) scanf("%d", a + i);
  sort(a + 1, a + n + 1);
  mxa = a[n] - a[1];
}
int f[maxn][maxn], g[maxn][maxn];
int calc(int Lim) {
  int i, j, k;
  for (i = 1; i <= K; i++) g[0][i] = 0;
  for (i = 1; i <= n; i++) f[i][1] = 1, g[i][1] = g[i - 1][1] + 1;
  for (k = 2; k <= K; k++) {
    j = 0;
    for (i = 1; i <= n; i++) {
      for (; a[i] - a[j + 1] >= Lim; j++)
        ;
      f[i][k] = g[j][k - 1];
      g[i][k] = MOD(g[i - 1][k] + f[i][k]);
    }
  }
  return g[n][K];
}
void solve() {
  int i, t, fans = 0;
  for (i = 1;; i++) {
    t = calc(i);
    if (!t) break;
    fans = MOD(fans + t);
  }
  printf("%d\n", fans);
}
int main() {
  input();
  solve();
  return 0;
}
