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 maxn = 1000 + 10;
const int MOD = 998244353;
const int DEBUG = 0;
inline int inc(int a, int b) {
  return (a + b >= MOD) ? (a + b - MOD) : (a + b);
}
inline int dec(int a, int b) { return (a >= b) ? (a - b) : (a + MOD - b); }
inline int mul(int a, int b) { return 1LL * a * b % MOD; }
inline int power(int x, int k) {
  int tmp = 1;
  while (k) {
    if (k & 1) tmp = mul(tmp, x);
    x = mul(x, x);
    k >>= 1;
  }
  return tmp;
}
int n, k, Ans, a[maxn], dp[maxn][maxn], sum[maxn][maxn];
inline void solve(int x) {
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= n; j++) dp[i][j] = sum[i][j] = 0;
  }
  for (int i = 1; i <= n; i++) dp[1][i] = 1;
  for (int i = 1; i <= n; i++) sum[1][i] = inc(sum[1][i - 1], dp[1][i]);
  for (int i = 2; i <= k; i++) {
    sum[i][0] = 0;
    int l = 1;
    for (int j = 1; j <= n; j++) {
      while (a[l] <= a[j] - x) l++;
      dp[i][j] = sum[i - 1][l - 1];
      sum[i][j] = inc(sum[i][j - 1], dp[i][j]);
    }
  }
  int now = 0;
  for (int i = 1; i <= n; i++) now = inc(now, dp[k][i]);
  Ans = inc(Ans, now);
  if (DEBUG) printf("%d %d\n", x, now);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  Ans = 0;
  for (int i = 1; i <= a[n] / (k - 1); i++) {
    solve(i);
  }
  printf("%d\n", Ans);
  return 0;
}
