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 mod = 998244353;
const int MAXN = 1e3 + 5;
int f[MAXN][MAXN], sum[MAXN][MAXN], a[MAXN], n, k, ans;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
inline void write(int x) {
  if (!x) {
    putchar('0');
    return;
  }
  char F[205];
  int tmp = x > 0 ? x : -x;
  if (x < 0) putchar('-');
  int cn = 0;
  while (tmp > 0) {
    F[cn++] = tmp % 10 + '0';
    tmp /= 10;
  }
  while (cn > 0) putchar(F[--cn]);
}
inline int mul(int x) {
  int res = 0;
  f[0][0] = sum[0][0] = 1;
  for (int i = 1, pt = 0; i <= n; i++) {
    while (pt + 1 <= i && a[i] - a[pt + 1] >= x) pt++;
    for (int j = 0; j <= k; j++) {
      if (j) f[i][j] = sum[pt][j - 1];
      sum[i][j] = (sum[i - 1][j] + f[i][j]) % mod;
    }
    res = (res + f[i][k]) % mod;
  }
  return res;
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  sort(a + 1, a + 1 + n);
  for (int i = (a[n] - a[1]) / (k - 1), last = 0; ~i; i--) {
    int now = mul(i);
    ans = (ans + 1ll * (now - last + mod) * i % mod) % mod;
    last = now;
  }
  write(ans);
  return 0;
}
