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;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
const int mod = 998244353;
const int maxn = 1010;
int n, m, k;
int a[maxn], ans[100100];
int f[1010][1010];
inline int sum(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int solve(int v) {
  for (register int i = (int)(0); i <= (int)(n); i++)
    for (register int j = (int)(0); j <= (int)(k); j++) f[i][j] = 0;
  int pos = 0;
  f[0][0] = 1;
  f[1][0] = 1;
  f[1][1] = 1;
  for (register int i = (int)(2); i <= (int)(n); i++) {
    while (a[i] - a[pos + 1] >= v) pos++;
    for (register int j = (int)(0); j <= (int)(k); j++) {
      if (j > 0) f[i][j] = sum(f[i][j], f[pos][j - 1]);
      f[i][j] = sum(f[i][j], f[i - 1][j]);
    }
  }
  return f[n][k];
}
void work() {
  n = read(), k = read();
  for (register int i = (int)(1); i <= (int)(n); i++) a[i] = read();
  sort(a + 1, a + n + 1);
  for (register int v = (int)(1); v <= (int)(a[n] / (k - 1)); v++) {
    ans[v] = solve(v);
    ans[v - 1] = (ans[v - 1] - ans[v] + mod) % mod;
  }
  int res = 0;
  for (register int v = (int)(1); v <= (int)(a[n] / (k - 1)); v++)
    res = sum(res, ans[v] * 1ll * v % mod);
  printf("%d\n", res);
}
int main() { work(); }
