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() {
  register char c = ' ';
  register int v = 0, x = 1;
  while ((c < '0') || (c > '9')) {
    if (c == '-') x = -1;
    c = getchar();
  }
  while ((c >= '0') && (c <= '9')) {
    v = (v << 1) + (v << 3) + (c ^ 48);
    c = getchar();
  }
  return v * x;
}
inline void write(register int x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 ^ 48);
}
const int N = 1010, M = 1e5 + 10, mod = 998244353;
int n, k, ans, a[N], res[M], dp[N][N], sum[N][N];
int main() {
  n = read(), k = read(), a[0] = -1e9;
  for (register int i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  for (register int v = 1; v * (k - 1) <= a[n]; v++) {
    register int now = 0;
    dp[0][0] = sum[0][0] = 1;
    for (register int i = 1; i <= n; i++) {
      while (a[now] <= a[i] - v) now++;
      for (register int j = 0; j <= k; j++) {
        if (j) dp[i][j] = sum[now - 1][j - 1];
        sum[i][j] = (sum[i - 1][j] + dp[i][j]) % mod;
      }
      res[v] = (res[v] + dp[i][k]) % mod;
    }
    ans = (ans + res[v]) % mod;
  }
  write(ans);
  return ~~(0 - 0);
}
