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;
template <typename T1, typename T2>
inline T1 max(T1 a, T2 b) {
  return a < b ? b : a;
}
template <typename T1, typename T2>
inline T1 min(T1 a, T2 b) {
  return a < b ? a : b;
}
namespace ae86 {
const int bufl = 1 << 15;
char buf[bufl], *s = buf, *t = buf;
inline int fetch() {
  if (s == t) {
    t = (s = buf) + fread(buf, 1, bufl, stdin);
    if (s == t) return EOF;
  }
  return *s++;
}
inline int ty() {
  int a = 0, b = 1, c = fetch();
  while (!isdigit(c)) b ^= c == '-', c = fetch();
  while (isdigit(c)) a = a * 10 + c - 48, c = fetch();
  return b ? a : -a;
}
}  // namespace ae86
using ae86::ty;
const int _ = 1007, mo = 998244353;
int n, len, val[_], f[_][_], sf[_][_];
inline int finder(int tar) {
  for (int i = 0; i <= len; i++)
    for (int j = 0; j <= n; j++) f[i][j] = sf[i][j] = 0;
  for (int i = 1; i <= n; i++) f[1][i] = 1;
  for (int i = 1; i <= len; i++)
    for (int j = 1, k = 0; j <= n; j++) {
      while (k < n && val[j] - val[k + 1] >= tar) k++;
      f[i][j] = (f[i][j] + sf[i - 1][k]) % mo,
      sf[i][j] = (sf[i][j - 1] + f[i][j]) % mo;
    }
  int ans = 0;
  for (int i = 1; i <= n; i++) ans = (ans + f[len][i]) % mo;
  return ans;
}
int main() {
  n = ty(), len = ty();
  for (int i = 1; i <= n; i++) val[i] = ty();
  sort(val + 1, val + n + 1);
  int ans = 0;
  for (int i = 1; i <= val[n] / (len - 1); i++) ans = (ans + finder(i)) % mo;
  printf("%d\n", ans);
  return 0;
}
