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;
long long read() {
  int f = 1;
  long long res = 0;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = -f;
  } while (ch < '0' || ch > '9');
  do {
    res = res * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  return f == 1 ? res : -res;
}
void fast_io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
const int N = 1010;
int mod = 998244353;
const long long INF = 1e18;
int n, m, q;
int A[N], B[N];
char str[N];
int head[N], to[N * 2], nxt[N * 2], tot = 1;
void addEdge(int u, int v, int val) {
  tot++;
  nxt[tot] = head[u];
  to[tot] = v;
  head[u] = tot;
}
template <class T>
T mmax(T a, T b) {
  return a < b ? b : a;
}
template <class T>
T mmin(T a, T b) {
  return a < b ? a : b;
}
int countOne(long long set) {
  int res = 0;
  while (set) {
    res++;
    set &= set - 1;
  }
  return res;
}
bool contain(int set, int i) { return (set & (1 << i)) > 0; }
long long myPow(long long a, int p) {
  if (p == 0) return 1;
  long long res = myPow(a, p / 2);
  res *= res;
  res %= mod;
  if (p % 2 == 1) {
    res *= a;
    res %= mod;
  }
  return res;
}
void addMode(long long &a, long long b) {
  a = (a + b) % mod;
  if (a < 0) a += mod;
}
long long mul(long long a, long long b) {
  long long res = a * b % mod;
  if (res < 0) res += mod;
  return res;
}
template <class T>
void mySwap(T &a, T &b) {
  T tmp = a;
  a = b;
  b = tmp;
}
int k;
int dp[N][N];
int le[N];
int pre[N][N];
int solve(int val) {
  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] >= val) le[i]++;
  }
  dp[0][0] = 1;
  for (int i = 0; i < k; i++) {
    pre[i][0] = dp[i][0];
    for (int j = 1; j <= n; j++) {
      pre[i][j] = (pre[i][j - 1] + dp[i][j]) % mod;
    }
    for (int j = 1; j <= n; j++) dp[i + 1][j] = pre[i][le[j]];
  }
  int res = 0;
  for (int i = 1; i <= n; i++) res = (res + dp[k][i]) % mod;
  return res;
}
int main() {
  fast_io();
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> A[i];
  sort(A + 1, A + n + 1);
  int ans = 0;
  for (int i = 1; i * (k - 1) <= 100010; i++) {
    ans = (ans + solve(i)) % mod;
  }
  cout << ans << endl;
  return 0;
}
