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>
#pragma GCC optimize(2)
using namespace std;
void read(long long &x) {
  x = 0;
  long long f = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -f;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  x *= f;
}
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (!x) return;
  print(x / 10), putchar(x % 10 + 48);
}
void write(long long x) {
  if (!x)
    putchar('0');
  else
    print(x);
  putchar('\n');
}
const long long maxn = 1e3 + 10;
const long long inf = 1e9;
const double eps = 1e-8;
const long long mod = 998244353;
long long n, k, f[maxn][maxn], a[maxn], t[maxn * 100], ans, g[maxn][maxn];
long long add(long long x, long long y) {
  return x + y >= mod ? x + y - mod : x + y;
}
void solve(long long x) {
  long long p = a[n] / x + 1;
  for (long long i = 1; i <= n; i++) {
    f[i][1] = 1, g[i][1] = i;
    if (a[i] < x) continue;
    long long r = t[a[i] - x], rr = min(i, p);
    for (long long j = 2; j <= rr; j++) {
      f[i][j] = g[r][j - 1];
      g[i][j] = add(g[i - 1][j], f[i][j]);
    }
  }
  ans = add(ans, g[n][k]);
  for (long long i = 1; i <= n; i++) {
    long long rr = min(i, p);
    for (long long j = 2; j <= rr; j++) g[i][j] = 0;
  }
}
signed main() {
  read(n), read(k);
  for (long long i = 1; i <= n; i++) read(a[i]);
  long long st = clock();
  sort(a + 1, a + n + 1);
  for (long long i = 1; i <= n - 1; i++)
    for (long long j = a[i]; j <= a[i + 1] - 1; j++) t[j] = i;
  for (long long i = 1; i * (k - 1) <= a[n] - a[1]; i++) solve(i);
  write(ans);
  cerr << (double)(clock() - st) / 1e3 << endl;
  return 0;
}
