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>
template <typename T>
inline int chkmin(T &x, const T &y) {
  return x > y ? x = y, 1 : 0;
}
template <typename T>
inline int chkmax(T &x, const T &y) {
  return x < y ? x = y, 1 : 0;
}
template <typename T>
inline T read(T &f) {
  f = 0;
  int x = 1;
  char c = getchar();
  while (!isdigit(c)) c = x == '-' ? -1 : 1, c = getchar();
  while (isdigit(c)) (f *= 10) += c & 15, c = getchar();
  return f *= x;
}
using int64 = long long;
using uint64 = unsigned long long;
const int N = 1000 + 5, KCZ = 998244353;
void add(int &x, const int &y) {
  x += y;
  if (x >= KCZ) x -= KCZ;
}
int n, k, A[N], L[N], S[N][N], f[N][N];
int dp(int mx) {
  for (int i = (0), ___ = (k); i <= ___; ++i)
    for (int j = (0), ___ = (n); j <= ___; ++j) f[i][j] = S[i][j] = 0;
  L[0] = 0;
  for (int i = (1), ___ = (n); i <= ___; ++i) {
    L[i] = L[i - 1];
    while (A[i] - A[L[i] + 1] >= mx) L[i]++;
  }
  f[0][0] = S[0][0] = 1;
  for (int i = (1), ___ = (k); i <= ___; ++i) {
    for (int j = (1), ___ = (n); j <= ___; ++j) {
      add(S[i - 1][j], f[i - 1][j]);
      add(S[i - 1][j], S[i - 1][j - 1]);
    }
    for (int j = (1), ___ = (n); j <= ___; ++j) {
      f[i][j] = S[i - 1][L[j]];
    }
  }
  int ret = 0;
  for (int i = (1), ___ = (n); i <= ___; ++i) add(ret, f[k][i]);
  return ret;
}
int main() {
  read(n), read(k);
  for (int i = (1), ___ = (n); i <= ___; ++i) read(A[i]);
  std::sort(A + 1, A + 1 + n);
  int Ans = 0;
  for (int mxA = (1), ___ = (100000); mxA <= ___; ++mxA) {
    if (mxA * (k - 1) > 100000) break;
    add(Ans, dp(mxA));
  }
  printf("%d\n", Ans);
  return 0;
}
