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 char gc() {
  static char now[1 << 16], *S, *T;
  if (S == T) {
    T = (S = now) + fread(now, 1, 1 << 16, stdin);
    if (S == T) return EOF;
  }
  return *S++;
}
inline int read() {
  int x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    (c == '-') ? (f = 0) : 0;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f ? x : ((~x) + 1);
}
const int mod = 998244353;
inline int add(const int &x, const int &y) {
  return (x + y >= mod) ? (x + y - mod) : (x + y);
}
inline void Add(int &x, const int &y) {
  x += y;
  (x >= mod) ? (x -= mod) : 0;
}
inline int sub(const int &x, const int &y) {
  return (x < y) ? (x - y + mod) : (x - y);
}
inline void Sub(int &x, const int &y) {
  x -= y;
  (x < 0) ? (x += mod) : 0;
}
inline int qpow(int a, int b) {
  if (!a) return 0;
  int res = 1;
  for (; b; b >>= 1, a = 1ll * a * a % mod)
    if (b & 1) res = 1ll * res * a % mod;
  return res;
}
const int N = 1010;
int n, m, a[N], f[N][N], g[N][N];
int main() {
  n = read();
  m = read();
  for (int(i) = (1); (i) <= (n); ++(i)) a[i] = read();
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int i = 1; i * (m - 1) <= 100000; ++i) {
    int k = 0;
    f[0][0] = g[0][0] = 1;
    for (int(j) = (1); (j) <= (n); ++(j)) {
      while (k + 1 < j && a[k + 1] + i <= a[j]) ++k;
      for (int(v) = (1); (v) <= (m); ++(v)) f[j][v] = add(f[j][v], g[k][v - 1]);
      for (int(v) = (0); (v) <= (m); ++(v)) g[j][v] = add(g[j - 1][v], f[j][v]);
    }
    ans = add(ans, g[n][m]);
    for (int(j) = (0); (j) <= (n); ++(j))
      for (int(k) = (0); (k) <= (m); ++(k)) f[j][k] = g[j][k] = 0;
  }
  printf("%d\n", ans);
  return 0;
}
