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;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x > y ? y : x;
}
const int INF = 2139062143;
template <typename T>
void read(T &x) {
  x = 0;
  bool f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = 0;
  } while (ch > '9' || ch < '0');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  x = f ? x : -x;
}
template <typename T>
void write(T x) {
  if (x < 0) x = ~x + 1, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 1000 + 5;
const int mod = 998244353;
int n, k, ans, a[N], f[N][N], s[N][N];
int main() {
  read(n);
  read(k);
  for (int i = 1; i <= n; i++) read(a[i]);
  sort(a + 1, a + n + 1);
  a[0] = -INF;
  for (int v = 1; v * (k - 1) <= a[n]; v++) {
    f[0][0] = s[0][0] = 1;
    int now = 0;
    for (int i = 1; i <= n; i++) {
      while (a[i] - a[now] >= v) now++;
      f[i][0] = 0;
      s[i][0] = (s[i - 1][0] + f[i][0]) % mod;
      for (int j = 1; j <= k; j++) {
        f[i][j] = s[now - 1][j - 1];
        s[i][j] = (s[i - 1][j] + f[i][j]) % mod;
      }
    }
    ans = (ans + 1ll * s[n][k]) % mod;
  }
  printf("%d\n", ans);
  return 0;
}
