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;
inline long long read() {
  long long res = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    res = (res << 3) + (res << 1) + ch - '0';
    ch = getchar();
  }
  return res * f;
}
void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 | '0');
}
const long long N = 1005;
const long long mo = 998244353;
long long ans[100005];
long long n, k, a[N], f[N][N];
long long Lim, Ans;
signed main() {
  n = read();
  k = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + n + 1);
  Lim = a[n] / (k - 1) + 1;
  for (long long v = 1; v <= Lim; v++) {
    for (long long i = 1; i <= n; i++)
      for (long long j = 2; j <= k; j++) f[i][j] = 0;
    for (long long i = 1; i <= n; i++) f[i][1] = 1;
    for (long long j = 2; j <= k; j++) {
      long long l = 0, sum = 0;
      for (long long i = 1; i <= n; i++) {
        while (a[i] - a[l + 1] >= v) {
          (sum += f[l + 1][j - 1]) %= mo;
          l++;
        }
        f[i][j] = sum;
      }
    }
    for (long long i = 1; i <= n; i++) (ans[v] += f[i][k]) %= mo;
  }
  for (long long i = Lim - 1; i >= 0; i--) {
    long long d = ((ans[i] - ans[i + 1]) % mo + mo) % mo;
    Ans = (Ans + d * i) % mo;
  }
  write((Ans % mo + mo) % mo);
  return 0;
}
