Kleofáš is participating in an n-thlon - a tournament consisting of n different competitions in n different disciplines (numbered 1 through n). There are m participants in the n-thlon and each of them participates in all competitions.

In each of these n competitions, the participants are given ranks from 1 to m in such a way that no two participants are given the same rank - in other words, the ranks in each competition form a permutation of numbers from 1 to m. The score of a participant in a competition is equal to his/her rank in it.

The overall score of each participant is computed as the sum of that participant's scores in all competitions.

The overall rank of each participant is equal to 1 + k, where k is the number of participants with strictly smaller overall score.

The n-thlon is over now, but the results haven't been published yet. Kleofáš still remembers his ranks in each particular competition; however, he doesn't remember anything about how well the other participants did. Therefore, Kleofáš would like to know his expected overall rank.

All competitors are equally good at each discipline, so all rankings (permutations of ranks of everyone except Kleofáš) in each competition are equiprobable.

Input

The first line of the input contains two space-separated integers n (1 ≤ n ≤ 100) and m (1 ≤ m ≤ 1000) — the number of competitions and the number of participants respectively.

Then, n lines follow. The i-th of them contains one integer xi (1 ≤ xi ≤ m) — the rank of Kleofáš in the i-th competition.

Output

Output a single real number – the expected overall rank of Kleofáš. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if <image>.

Examples

Input

4 10
2
1
2
1


Output

1.0000000000000000


Input

5 5
1
2
3
4
5


Output

2.7500000000000000


Input

3 6
2
4
2


Output

1.6799999999999999

Note

In the first sample, Kleofáš has overall score 6. Nobody else can have overall score less than 6 (but it's possible for one other person to have overall score 6 as well), so his overall rank must be 1.
#include <bits/stdc++.h>
namespace IO {
const int SIZE = (1 << 20) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = obuf + SIZE - 1;
char _st[55];
int _qr = 0;
inline char gc() {
  return (iS == iT ? iT = (iS = ibuf) + fread(ibuf, 1, SIZE, stdin),
          (iS == iT ? EOF : *iS++) : *iS++);
}
inline void qread() {}
template <class T1, class... T2>
inline void qread(T1 &IEE, T2 &...ls) {
  register T1 __ = 0, ___ = 1;
  register char ch;
  while (!isdigit(ch = gc())) ___ = (ch == '-') ? -___ : ___;
  do {
    __ = (__ << 1) + (__ << 3) + (ch ^ 48);
  } while (isdigit(ch = gc()));
  __ *= ___;
  IEE = __;
  qread(ls...);
  return;
}
template <class T>
inline void QreadArr(T Begin, T End) {
  while (Begin != End) {
    qread(*Begin);
    ++Begin;
  }
}
inline void flush() {
  fwrite(obuf, 1, oS - obuf, stdout);
  oS = obuf;
  return;
}
inline void putc_(char _x) {
  *oS++ = _x;
  if (oS == oT) flush();
}
inline void qwrite() {}
template <class T1, class... T2>
inline void qwrite(T1 IEE, T2... ls) {
  if (!IEE) putc_('0');
  if (IEE < 0) putc_('-'), IEE = -IEE;
  while (IEE) _st[++_qr] = IEE % 10 + '0', IEE /= 10;
  while (_qr) putc_(_st[_qr--]);
  qwrite(ls...);
  return;
}
template <class T>
inline void WriteArr(T Begin, T End) {
  const char Kg = ' ', Edl = '\n';
  while (Begin != End) {
    qwrite(*Begin);
    ++Begin;
    putc_(Kg);
  }
  putc_(Edl);
}
struct Flusher_ {
  ~Flusher_() { flush(); }
} io_flusher;
}  // namespace IO
using namespace IO;
using namespace std;
mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long My_Rand(long long Mod) { return (unsigned long long)(rnd()) % Mod; }
template <class T1, class T2>
void To_max(T1 &x, T2 y) {
  x < y ? x = y : x = x;
}
template <class T1, class T2>
void To_min(T1 &x, T2 y) {
  x > y ? x = y : x = x;
}
inline long long qpow(long long n, long long base, long long mod = 1e18) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * base % mod;
    base = base * base % mod;
    n >>= 1;
  }
  return ret % mod;
}
inline long long gcd(long long x, long long y) {
  return !y ? x : gcd(y, x % y);
}
inline long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
const int N = 1005;
const int M = 1e5 + 5;
int n, m, Sum;
int c[N];
double ans;
double dp[2][M];
int main() {
  dp[0][0] = 1.0;
  qread(n, m);
  QreadArr(c + 1, c + n + 1);
  for (int i = 1; i <= n; ++i) Sum += c[i];
  int lst = 0, now = 0;
  for (int i = 1; i <= n; ++i) {
    lst = now, now ^= 1;
    double sum = dp[lst][0];
    for (int j = 1; j <= Sum - 1; ++j) {
      dp[now][j] = sum;
      if (j >= c[i]) dp[now][j] -= dp[lst][j - c[i]];
      sum += dp[lst][j];
      if (j >= m) sum -= dp[lst][j - m];
      dp[now][j] /= (m - 1);
    }
    if (i == 1) dp[0][0] = 0.0;
  }
  for (int i = n; i <= Sum - 1; ++i) ans += dp[now][i];
  printf("%.15lf\n", ans * (m - 1) + 1.0);
  return 0;
}
