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 N = 1010, mod = 998244353;
inline int read() {
  register int x = 0;
  register char c = getchar();
  while (c < '0' || c > '9') c = getchar();
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x;
}
int n, k;
int a[N];
inline int Count(int x) {
  int ans = 0;
  while (x) ans += x & 1, x >>= 1;
  return ans;
}
inline int Abs(int x) { return x < 0 ? -x : x; }
inline int add(int a, int b) {
  a += b;
  return a >= mod ? a - mod : a;
}
inline int mul(int a, int b) { return (long long)a * b % mod; }
inline int ksm(int a, int k) {
  int ans = 1;
  a %= mod;
  for (; k; k >>= 1, a = mul(a, a))
    if (k & 1) ans = mul(ans, a);
  return ans;
}
namespace Work1 {
int b[20];
int main() {
  long long ans = 0;
  for (int i = 1; i < (1 << n); ++i)
    if (Count(i) == k) {
      int tot = 0, mn = 2e9;
      for (int j = 1; j <= n; ++j)
        if ((i >> (j - 1)) & 1) b[++tot] = a[j];
      for (int j = 1; j < tot; ++j) mn = min(mn, Abs(b[j] - b[j + 1]));
      (ans += mn) %= mod;
    }
  cout << ans << '\n';
  return 0;
}
}  // namespace Work1
namespace Work2 {
int fac[N], inv[N];
void Init() {
  fac[0] = inv[0] = 1;
  for (int i = 1; i <= n; ++i) fac[i] = mul(fac[i - 1], i);
  inv[n] = ksm(fac[n], mod - 2);
  for (int i = n - 1; i >= 1; --i) inv[i] = mul(inv[i + 1], i + 1);
}
inline int C(int n, int m) {
  return n >= m ? mul(fac[n], mul(inv[m], inv[n - m])) : 0;
}
int main() {
  if (k < 2) return puts("0"), 0;
  long long ans = 0;
  Init();
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j) {
      int d = Abs(a[i] - a[j]);
      int l = lower_bound(a + 1, a + 1 + n, a[i] - d) - a - 1;
      int r = upper_bound(a + 1, a + 1 + n, a[j] + d) - a;
      int cnt = l + n - r + 1;
      ans = add(ans, mul(d, C(cnt, k - 2)));
    }
  cout << ans << '\n';
  return 0;
}
}  // namespace Work2
namespace Work3 {
int f[1010][1010], s[1010][1010];
int main() {
  int ans = 0;
  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[now + 1] <= a[i] - v) now++;
      for (int j = 0; j <= k; ++j) {
        if (j) f[i][j] = s[now][j - 1];
        s[i][j] = add(s[i - 1][j], f[i][j]);
      }
      ans = add(ans, f[i][k]);
    }
  }
  cout << ans << '\n';
  return 0;
}
}  // namespace Work3
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  sort(a + 1, a + 1 + n);
  Work3::main();
  fclose(stdin);
  fclose(stdout);
  return 0;
}
