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 io {
const int SI = 1 << 21 | 1;
char IB[SI], *IS, *IT, OB[SI], *OS = OB, *OT = OS + SI - 1, c, ch[100];
int f, t;
inline void flush() { fwrite(OB, 1, OS - OB, stdout), OS = OB; }
inline void pc(char x) {
  *OS++ = x;
  if (OS == OT) flush();
}
template <class I>
inline void rd(I &x) {
  for (f = 1, c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                              IS == IT ? EOF : *IS++)
                            : *IS++);
       c < '0' || c > '9';
       c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                       IS == IT ? EOF : *IS++)
                     : *IS++))
    if (c == '-') f = -1;
  for (x = 0; c >= '0' && c <= '9'; x = (x << 3) + (x << 1) + (c & 15),
      c = (IS == IT ? (IT = (IS = IB) + fread(IB, 1, SI, stdin),
                      IS == IT ? EOF : *IS++)
                    : *IS++))
    ;
  x *= f;
}
template <class I>
inline void print(I x, char k = '\n') {
  if (!x) pc('0');
  if (x < 0) pc('-'), x = -x;
  while (x) ch[++t] = x % 10 + '0', x /= 10;
  while (t) pc(ch[t--]);
  pc(k);
}
struct Flush {
  ~Flush() { flush(); }
} flusher;
}  // namespace io
using io::print;
using io::rd;
template <class I>
inline I Min(I x, I y) {
  return x < y ? x : y;
}
template <class I>
inline I Max(I x, I y) {
  return x > y ? x : y;
}
template <class I>
inline void Swap(I &x, I &y) {
  x ^= y ^= x ^= y;
}
const int N = 1e3 + 6, V = 1e5, P = 998244353;
int n, m, s[N], ans[V + 1], f[N][N], g[N][N];
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> s[i];
  sort(s + 1, s + n + 1);
  s[0] = -V - 1;
  int ans = 0;
  for (int i = 1; i * (m - 1) <= V; i++) {
    int k = 0;
    g[0][0] = f[0][0] = 1;
    for (int j = 1; j <= n; j++) {
      for (; k + 1 < j && s[k + 1] + i <= s[j]; k++)
        ;
      for (int l = 1; l <= m; l++) (f[j][l] += g[k][l - 1]) %= P;
      for (int l = 0; l <= m; l++) g[j][l] = (g[j - 1][l] + f[j][l]) % P;
    }
    (ans += g[n][m]) %= P;
    for (int j = 0; j <= n; j++)
      for (int k = 0; k <= m; k++) f[j][k] = g[j][k] = 0;
  }
  cout << ans << endl;
  return 0;
}
