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 N = 1010;
int n, k, lmt, a[N], ans = 0;
int f[N][N], sum[N], fr[N], cnt[100010];
inline int add(const int &x, const int &y) {
  return x + y < mod ? x + y : x + y - mod;
}
inline int sub(const int &x, const int &y) {
  return x - y < 0 ? x - y + mod : x - y;
}
inline int mul(const int &x, const int &y) {
  return (int)((long long)x * y % mod);
}
int ksm(int x, int y = mod - 2) {
  int ss = 1;
  for (; y; y >>= 1, x = mul(x, x))
    if (y & 1) ss = mul(ss, x);
  return ss;
}
int solve(int x) {
  f[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    fr[i] = fr[i - 1];
    for (; a[i] - a[fr[i] + 1] >= x; ++fr[i])
      ;
  }
  for (int j = 1; j <= k; j++) {
    sum[0] = f[0][j - 1];
    for (int i = 1; i <= n; i++) sum[i] = add(sum[i - 1], f[i][j - 1]);
    for (int i = 1; i <= n; i++) f[i][j] = sum[fr[i]];
  }
  for (int i = 0; i < n; i++) f[n][k] = add(f[n][k], f[i][k]);
  return f[n][k];
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1), lmt = a[n] / (k - 1) + 1;
  for (int i = 1; i <= lmt; i++) cnt[i] = solve(i);
  for (int i = 1; i <= lmt; i++)
    ans = add(ans, mul(i, sub(cnt[i], cnt[i + 1])));
  printf("%d\n", ans);
  return 0;
}
