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 int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 998244353;
template <typename S, typename T>
inline bool Min(S &a, const T &b) {
  return a > b ? a = b, true : false;
}
template <typename S, typename T>
inline bool Max(S &a, const T &b) {
  return a < b ? a = b, true : false;
}
template <typename S, typename T>
inline void Adm(S &a, const T &b) {
  a = (a + b) % MOD;
  if (a < 0) a += MOD;
}
template <typename S, typename T>
inline void Mum(S &a, const T &b) {
  a = a * b % MOD;
}
inline long long Mum(initializer_list<long long> ls) {
  long long a = 1;
  for (long long b : ls) a = a * b % MOD;
  return a;
}
const int N = 1e3 + 100;
int a[N], n, k;
long long f[N][N], g[100010];
long long dp(int x) {
  for (int i = 0; i <= n + 5; ++i)
    for (int j = 0; j <= k; ++j) f[i][j] = 0;
  f[0][0] = 1;
  int p = 0;
  for (int i = 1; i <= n; ++i) {
    while (p + 1 < i && a[i] - a[p + 1] >= x) ++p;
    for (int j = 1; j <= k; ++j) Adm(f[i][j], f[p][j - 1]);
    for (int j = 0; j <= k; ++j) Adm(f[i][j], f[i - 1][j]);
  }
  return f[n][k];
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  long long ans = 0;
  for (int i = 100000 / (k - 1); i >= 1; --i) {
    g[i] = dp(i);
    Adm(ans, (g[i] - g[i + 1]) * i);
  }
  cout << ans << endl;
  return 0;
}
