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("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target( \
    "sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,avx2,tune=native")
using namespace std;
template <typename T1, typename T2>
inline void mine(T1 &x, const T2 &y) {
  if (y < x) x = y;
}
template <typename T1, typename T2>
inline void maxe(T1 &x, const T2 &y) {
  if (x < y) x = y;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &b) {
  for (auto &k : b) os << k << ' ';
  return os;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &b) {
  for (auto &k : b) is >> k;
  return is;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 998244353;
namespace md {
int mod(long long n) {
  if (n <= -MOD || n >= MOD) n %= MOD;
  if (n < 0) return n + MOD;
  return n;
}
int add(int a, int b) {
  a += b;
  if (a >= MOD) return a - MOD;
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) return a + MOD;
  return a;
}
long long mult(long long a, long long b) { return a * b % MOD; }
long long powmod(long long a, long long p) {
  if (p == 0) return 1;
  if (p & 1) return mult(a, powmod(a, p - 1));
  return powmod(mult(a, a), p / 2);
}
long long rev(long long a) { return powmod(a, MOD - 2); }
}  // namespace md
struct M {
  int x;
  M(long long _x) { x = md::mod(_x); }
  M() { x = 0; }
  M operator+(M y) { return M(md::add(x, y.x)); }
  M operator-(M y) { return M(md::sub(x, y.x)); }
  M operator*(M y) { return M(1ll * x * y.x); }
  M operator/(M y) { return M(x * md::rev(y.x)); }
  M operator+(int y) { return (*this) + M(y); }
  M operator-(int y) { return (*this) - M(y); }
  M operator*(int y) { return (*this) * M(y); }
  M operator/(int y) { return (*this) / M(y); }
  M operator^(int y) { return M(x ^ y); }
  void operator+=(M y) { *this = *this + y; }
  void operator-=(M y) { *this = *this - y; }
  void operator*=(M y) { *this = *this * y; }
  void operator/=(M y) { *this = *this / y; }
  void operator+=(int y) { *this = *this + y; }
  void operator-=(int y) { *this = *this - y; }
  void operator*=(int y) { *this = *this * y; }
  void operator/=(int y) { *this = *this / y; }
  void operator^=(int y) { *this = *this ^ y; }
  bool operator==(M y) { return x == y.x; }
  bool operator<(M y) { return x < y.x; }
};
ostream &operator<<(ostream &os, const M &a) {
  os << a.x;
  return os;
}
istream &operator>>(istream &is, M &a) {
  long long x;
  is >> x;
  a = M(x);
  return is;
}
const int N = 1001;
const int C = 1e5 + 10;
int a[N];
M dp[N][N];
M ps[N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a, a + n);
  M ans(0);
  for (int can = C; can > 0; --can) {
    int mxj = a[n - 1] / can + 1;
    if (mxj < k) continue;
    mine(mxj, k);
    int ptr = 0;
    for (int j = 0; j <= mxj; ++j) ps[j] = 0;
    for (int i = 0; i < n; ++i) {
      while (ptr < i && a[ptr] <= a[i] - can) {
        for (int j = 0; j <= mxj; ++j) {
          ps[j] += dp[ptr][j];
        }
        ++ptr;
      }
      for (int j = 1; j <= mxj; ++j) dp[i][j] = ps[j - 1];
      dp[i][1] += 1;
      ans += dp[i][k];
    }
  }
  cout << ans << '\n';
}
