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 maxn = 1e5 + 10;
long long n, k, x[maxn], ans, a[maxn], tot, vis[maxn];
long long f[1010][1010], s[1010][1010], res[maxn];
inline long long read() {
  long long x = 0, k = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') k = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + (c ^ 48);
    c = getchar();
  }
  return x * k;
}
inline int jc(int x) { return x == 1 || x == 0 ? 1 : x * jc(x - 1); }
inline int C(int n, int m) {
  if (n < m) swap(n, m);
  return jc(n) / (jc(n - m) * jc(m));
}
inline void check() {
  tot = 0;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 2) a[++tot] = i;
  }
  long long minn = (1 << 30);
  for (int i = 1; i <= k; i++)
    for (int j = i + 1; j <= k; j++) minn = min(minn, abs(x[a[j]] - x[a[i]]));
  ans = (ans + minn) % 998244353;
}
inline void dfs(int p, int cnt) {
  if (p == n + 1) {
    if (cnt == k) check();
    return;
  }
  for (int i = 1; i <= 2; i++) {
    if (!vis[p]) {
      vis[p] = i;
      if (i == 1)
        dfs(p + 1, cnt);
      else
        dfs(p + 1, cnt + 1);
      vis[p] = 0;
    }
  }
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; ++i) x[i] = read();
  if (n <= 10)
    dfs(1, 0);
  else {
    sort(x + 1, x + n + 1);
    x[0] = -(1 << 30);
    for (int v = 1; v * (k - 1) <= x[n]; v++) {
      f[0][0] = 1;
      int now = 0;
      s[0][0] = 1;
      for (int i = 1; i <= n; i++) {
        while (x[now] <= x[i] - v) now++;
        for (int j = 0; j <= k; j++) {
          if (j) f[i][j] = s[now - 1][j - 1];
          s[i][j] = (s[i - 1][j] + f[i][j]) % 998244353;
        }
        res[v] += f[i][k];
      }
      ans += res[v] % 998244353;
    }
  }
  printf("%lld", ans % 998244353);
  return 0;
}
