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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int N = 1007, mo = 998244353;
inline int fk(int x) { return x >= mo ? x - mo : x; }
int n, m, A[N];
int f[N][N], ans;
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) A[i] = read();
  sort(A + 1, A + n + 1);
  int mx = A[n];
  for (int I = 1; I <= mx / (m - 1); I++) {
    for (int i = 1; i <= n; i++) f[1][i] = 1;
    for (int i = 2; i <= m; i++) {
      int sum = 0, p = 1;
      for (int j = 1; j <= n; j++) {
        while (A[p] + I <= A[j]) sum = fk(sum + f[i - 1][p++]);
        f[i][j] = sum;
      }
    }
    for (int i = 1; i <= n; i++) ans = fk(ans + f[m][i]);
  }
  printf("%d\n", ans);
  return 0;
}
