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;
long long read() {
  long long w = 0, ff = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    w = w * 10 + ch - '0';
    ch = getchar();
  }
  return ff * w;
}
const long long P = 998244353;
long long ans, dp[1005][1005], a[1005], n, k, now;
long long get(long long val) {
  dp[0][0] = 1, now = 0;
  for (long long i = 1; i <= n; ++i) {
    while (a[i] - a[now + 1] >= val) now++;
    dp[i][0] = 1;
    for (long long j = 1; j <= k; ++j)
      dp[i][j] = (dp[now][j - 1] + dp[i - 1][j]) % P;
  }
  return dp[n][k];
}
signed main() {
  n = read(), k = read();
  for (long long i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + n + 1);
  a[0] = -1e9;
  for (long long i = 1; (k - 1) * i <= a[n]; ++i) ans = (ans + get(i)) % P;
  printf("%lld", ans);
  return 0;
}
