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 = 3e5 + 5;
const int mod = 998244353;
int sum(int a, int b) {
  int s = (a + b);
  if (s >= mod) s -= mod;
  return s;
}
int sub(int a, int b) {
  int s = a - b;
  if (s < 0) s += mod;
  return s;
}
int mult(int a, int b) { return (1LL * a * b) % mod; }
int dp[1010][1010], pref[1010][1010], le[1010], a[1010];
int n, k, ans = 0;
long long rd() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int solve(int x) {
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= n; j++) dp[i][j] = 0;
  le[0] = 0;
  for (int i = 1; i <= n; i++) {
    le[i] = le[i - 1];
    while (a[i] - a[le[i] + 1] >= x) le[i]++;
  }
  dp[0][0] = 1;
  for (int i = 0; i <= k - 1; i++) {
    pref[i][0] = dp[i][0];
    for (int j = 1; j <= n; j++) pref[i][j] = sum(pref[i][j - 1], dp[i][j]);
    for (int j = 1; j <= n; j++) dp[i + 1][j] = pref[i][le[j]];
  }
  int ans = 0;
  for (int j = 1; j <= n; j++) ans = sum(ans, dp[k][j]);
  return ans;
}
int main() {
  n = rd();
  k = rd();
  for (int i = 1; i <= n; i++) a[i] = rd();
  sort(a + 1, a + 1 + n);
  for (int x = 1; (k - 1) * x <= 101010; x++) ans = sum(ans, solve(x));
  printf("%d\n", ans);
}
