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 = 1000 + 77, NN = 100000 + 77, Mod = 998244353;
int n, k, ans, A[NN], dp[N][N], ps[N][N], a[N], R[N];
int lim;
inline void Solve(int x) {
  R[n] = n + 1;
  for (int i = n - 1; i > 0; --i) {
    R[i] = R[i + 1];
    while (a[R[i] - 1] - a[i] >= x && R[i] > i + 1) --R[i];
  }
  dp[n + 1][0] = 1;
  ps[n + 1][0] = 1;
  for (int i = n; i > 0; --i) {
    ps[i][0] = 1;
    for (int j = 1; j <= k; ++j) {
      dp[i][j] = ps[R[i]][j - 1];
      ps[i][j] = (ps[i + 1][j] + dp[i][j]);
      if (ps[i][j] >= Mod) ps[i][j] -= Mod;
    }
    A[x] = (A[x] + dp[i][k]);
    if (A[x] >= Mod) A[x] -= 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);
  a[n + 1] = 1e9;
  int lim = (a[n] - a[1] + k) / (k - 1);
  for (int i = 0; i <= lim; ++i) Solve(i);
  int sum = 0;
  for (int i = lim; i >= 0; --i) {
    int now = (A[i] - sum + Mod) % Mod;
    ans = (ans + 1ll * now * i) % Mod;
    sum = (sum + now) % Mod;
  }
  printf("%d\n", ans);
  return 0;
}
