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;
namespace modular {
const int mod = 998244353;
inline int add(int a, int b) {
  a += b;
  while (a >= mod) a -= mod;
  while (a < 0) a += mod;
  return a;
}
inline void add_self(int& a, int b) { a = add(a, b); }
inline int mul(int a, int b) { return 1LL * a * b % mod; }
inline void mul_self(int& a, int b) { a = mul(a, b); }
inline int mpow(int a, int n) {
  int res = 1;
  while (n > 0) {
    if (n & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    n >>= 1;
  }
  return res;
}
inline int divide(int a, int b) {
  assert(b != 0);
  return mul(a, mpow(b, mod - 2));
}
const int mxN = 200 * 1000 + 5;
int fact[mxN], finv[mxN];
inline int binomial(int n, int k) {
  return mul(fact[n], mul(finv[n - k], finv[k]));
}
inline void init_fact() {
  fact[0] = finv[0] = 1;
  for (int i = 1; i < mxN; i++) {
    fact[i] = mul(i, fact[i - 1]);
    finv[i] = mpow(fact[i], mod - 2);
  }
}
}  // namespace modular
using namespace modular;
const int nax = 1005;
int a[nax], dp[nax][nax], n, k;
int solve_one(int B) {
  dp[0][0] = 1;
  int pos = 0;
  for (int i = 0; i < int(n); i++) {
    while (a[i] - a[pos] >= B) {
      pos++;
    }
    for (int j = 0; j < int(k + 1); j++) {
      dp[i + 1][j] = dp[i][j];
    }
    for (int j = 0; j < int(k); j++) {
      add_self(dp[i + 1][j + 1], dp[pos][j]);
    }
  }
  return dp[n][k];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  int A = 0;
  for (int i = 0; i < int(n); i++) {
    cin >> a[i];
    A = max(A, a[i]);
  }
  sort(a, a + n);
  int ans = 0;
  for (int b = 1; b <= A / (k - 1) + 1; b++) {
    add_self(ans, solve_one(b));
  }
  cout << ans << endl;
}
