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;
const int p = 998244353;
const int N = 1005;
const int MAX = 1e5 + 5;
template <class T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = x * 10 + s - '0';
    s = getchar();
  }
  x *= f;
}
int n, k, a[N], ans, f[N][N], mx, as[MAX], pos;
inline int calc(int x) {
  f[0][0] = 1;
  f[1][1] = f[1][0] = 1;
  pos = 0;
  for (int i = (2); i <= (n); i++) {
    while ((a[i] - a[pos + 1]) >= x) pos++;
    f[i][0] = 1;
    for (int j = (1); j <= (k); j++)
      f[i][j] = (f[i - 1][j] + f[pos][j - 1]) % p;
  }
  return f[n][k];
}
inline void solve() {
  read(n);
  read(k);
  for (int i = (1); i <= (n); i++) read(a[i]);
  sort(a + 1, a + n + 1);
  mx = (a[n] - a[1]) / (k - 1);
  for (int i = (1); i <= (mx); i++) as[i] = calc(i);
  for (int i = (1); i <= (mx); i++) as[i] -= as[i + 1], (as[i] += p) %= p;
  for (int i = (1); i <= (mx); i++) (ans += 1ll * i * as[i] % p) %= p;
  printf("%d\n", ans);
}
int main() { solve(); }
