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 = 1000 + 10, M = 1e5 + 10, mod = 998244353;
const double eps = 1e-5;
int rd() {
  int x = 0, w = 1;
  char ch = 0;
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + (ch ^ 48);
    ch = getchar();
  }
  return x * w;
}
void ad(int &x, int y) { x += y, x -= x >= mod ? mod : 0; }
int n, kk, a[N], nm[M], f[2][N];
int main() {
  n = rd(), kk = rd();
  for (int i = 1; i <= n; ++i) a[i] = rd();
  sort(a + 1, a + n + 1);
  for (int i = n; i; --i) a[i] -= a[1];
  a[0] = -(1 << 30);
  for (int h = 0; h <= a[n] / (kk - 1); ++h) {
    int nw = 1, la = 0;
    memset(f[la], 0, sizeof(f[la]));
    f[la][0] = 1;
    for (int i = 1; i <= kk; ++i) {
      memset(f[nw], 0, sizeof(f[nw]));
      int pr = 0;
      for (int j = 1, z = 0; j <= n; ++j) {
        while (z < j && a[j] - a[z] >= h) ad(pr, f[la][z]), ++z;
        f[nw][j] = pr;
      }
      nw ^= 1, la ^= 1;
    }
    for (int j = 1; j <= n; ++j) ad(nm[h], f[la][j]);
    if (!nm[h]) break;
  }
  int ans = 0;
  for (int h = 0; h <= a[n] / (kk - 1); ++h)
    ad(ans, 1ll * (nm[h] - nm[h + 1] + mod) * h % mod);
  printf("%d\n", ans);
  return 0;
}
