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 out = 0, fh = 1;
  char jp = getchar();
  while ((jp > '9' || jp < '0') && jp != '-') jp = getchar();
  if (jp == '-') fh = -1, jp = getchar();
  while (jp >= '0' && jp <= '9') out = out * 10 + jp - '0', jp = getchar();
  return out * fh;
}
const int P = 998244353;
int add(int a, int b) { return (a + b >= P) ? (a + b - P) : (a + b); }
const int MAXN = 1e3 + 10;
int n, k;
int f[MAXN][MAXN], sum[MAXN][MAXN], a[MAXN];
int solve(int x) {
  for (int i = 1; i <= n; ++i) f[i][1] = 1, sum[i][1] = i;
  for (int j = 2; j <= k; ++j) {
    sum[0][j] = 0;
    int r = 1;
    for (int i = 1; i <= n; ++i) {
      while (a[r] <= a[i] - x) ++r;
      f[i][j] = sum[r - 1][j - 1];
      sum[i][j] = add(sum[i - 1][j], f[i][j]);
    }
  }
  return sum[n][k];
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + 1 + n);
  int m = a[n];
  int ans = 0;
  for (int x = 1; x <= m / (k - 1); ++x) ans = add(ans, solve(x));
  cout << ans << endl;
  return 0;
}
