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 long long read() {
  long long sum = 0, nega = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') sum = sum * 10 + ch - '0', ch = getchar();
  return sum * nega;
}
const long long mod = 998244353;
long long n, k, a[1009], maxv, res[100009], ans, dp[1009][1009];
long long cnt[100009];
signed main() {
  n = read(), k = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  maxv = (a[n] - a[1]) / (k - 1);
  for (long long i = 1; i <= n; i++) cnt[a[i]] = i;
  for (long long i = 1; i <= a[n]; i++)
    if (!cnt[i]) cnt[i] = cnt[i - 1];
  for (long long val = maxv; val >= 0; val--) {
    for (long long i = 1; i <= n; i++)
      for (long long j = 1; j <= k; j++) dp[i][j] = 0;
    for (long long i = 1; i <= n; i++) dp[i][1] = i;
    for (long long i = 1; i <= n; i++) {
      long long to = 0;
      if (a[i] >= val) to = cnt[a[i] - val];
      if (to == i) to = i - 1;
      for (long long j = 2; j <= k && j <= i; j++) {
        dp[i][j] = (dp[i][j] + dp[to][j - 1]) % mod;
        dp[i][j] = (dp[i][j] + dp[i - 1][j]) % mod;
      }
    }
    res[val] = dp[n][k] % mod;
  }
  for (long long val = maxv; val >= 0; val--)
    ans = (ans + (res[val] - res[val + 1] + mod) * val % mod) % mod;
  cout << ans << endl;
  return 0;
}
