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 long long N = 1010, Mod = 998244353;
inline long long read() {
  long long s = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    s = (s << 3) + (s << 1) + (ch - '0'), ch = getchar();
  return s * w;
}
void print(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
long long n, K, a[N], g[N], dp[N][N], qz[N][N], pos[N], res;
signed main() {
  n = read(), K = read();
  for (register long long i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + 1 + n);
  a[0] = -1e15;
  long long qwq = a[n] - a[1];
  for (register long long i = 1; i <= qwq / (K - 1); i++) {
    dp[0][0] = qz[0][0] = 1;
    long long L = 1;
    for (register long long j = 1; j <= n; j++) {
      while (a[j] - a[L] >= i) L++;
      pos[j] = L - 1;
      qz[j][0] = qz[j - 1][0];
      for (register long long k = 1; k <= K; k++)
        dp[j][k] = qz[pos[j]][k - 1],
        qz[j][k] = (qz[j - 1][k] + dp[j][k]) % Mod;
      if (j >= K) res += dp[j][K], res %= Mod;
    }
  }
  printf("%lld\n", res);
  return 0;
}
