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 T>
void read(T &x) {
  x = 0;
  bool f = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = 1;
  for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
  if (f) x = -x;
}
template <typename F>
inline void write(F x, char ed = '\n') {
  static short st[30];
  short tp = 0;
  if (x < 0) putchar('-'), x = -x;
  do st[++tp] = x % 10, x /= 10;
  while (x);
  while (tp) putchar('0' | st[tp--]);
  putchar(ed);
}
template <typename T>
inline void Mx(T &x, T y) {
  x < y && (x = y);
}
template <typename T>
inline void Mn(T &x, T y) {
  x > y && (x = y);
}
const int P = 998244353;
const int N = 1005;
int f[N][N], a[N], mx, n, k;
inline int add(int x, int y) {
  x += y;
  return x >= P ? x - P : x;
}
int work(int lim) {
  for (int j = 2; j <= k; j++) {
    int l = j - 1, sum = 0, *A = f[j - 1], *B = f[j];
    for (int i = j; i <= n; i++) {
      while (a[i] - a[l] >= lim) sum = add(sum, A[l]), l++;
      B[i] = sum;
    }
  }
  int sum = 0;
  for (int i = k; i <= n; i++) sum = add(sum, f[k][i]);
  return sum;
}
int main() {
  read(n), read(k);
  for (int i = 1; i <= n; i++) read(a[i]), Mx(mx, a[i]);
  sort(a + 1, a + n + 1), mx /= (k - 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) f[1][i] = 1;
  for (int i = 1; i <= mx; i++) ans = add(ans, work(i));
  write(ans);
  return 0;
}
