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>
#pragma GCC optimize(3)
#pragma GCC optimize(2)
#pragma GCC optimize("Ofast")
using namespace std;
inline long long read() {
  long long num = 0;
  char c = ' ';
  for (; c > '9' || c < '0'; c = getchar())
    ;
  for (; c >= '0' && c <= '9';
       num = (num << 1) + (num << 3) + c - 48, c = getchar())
    ;
  return num;
}
long long n, k, a[1005];
long long dp[1005][1005];
long long sum[1005][1005];
inline long long solve(long long v) {
  dp[0][0] = 1, sum[0][0] = 1;
  long long pos = 0;
  for (long long i = 1; i <= n; i++) {
    while (a[i] - a[pos] >= v) pos++;
    for (long long j = 0; j <= k; j++) {
      if (j) dp[i][j] = sum[pos - 1][j - 1];
      sum[i][j] = (sum[i - 1][j] + dp[i][j]) % 998244353;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) ans = (ans + dp[i][k]) % 998244353;
  return ans;
}
signed main() {
  n = read(), k = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  a[0] = -1000000000ll;
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  for (long long x = 1; x * (k - 1) <= a[n]; x++)
    ans = (ans + solve(x)) % 998244353;
  cout << ans << endl;
  return 0;
}
