Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct base {
  double x, y;
  base() { x = y = 0; }
  base(double _x, double _y) { x = _x, y = _y; }
  base(double _x) { x = _x, y = 0; }
  void operator=(double _x) { x = _x, y = 0; }
  double real() { return x; }
  double imag() { return y; }
  base operator+(const base& b) { return base(x + b.x, y + b.y); }
  void operator+=(const base& b) { x += b.x, y += b.y; }
  base operator*(const base& b) {
    return base(x * b.x - y * b.y, x * b.y + y * b.x);
  }
  void operator*=(const base& b) {
    double p = x * b.x - y * b.y, q = x * b.y + y * b.x;
    x = p, y = q;
  }
  void operator/=(double k) { x /= k, y /= k; }
  base operator-(const base& b) { return base(x - b.x, y - b.y); }
  void operator-=(const base& b) { x -= b.x, y -= b.y; }
  base conj() { return base(x, -y); }
};
double PI = 2.0 * acos(0.0);
base w[1 << 20];
void fft(vector<base>& a, bool invert) {
  int n = (int)a.size();
  for (int i = 1, j = 0; i < n; ++i) {
    int bit = n >> 1;
    for (; j >= bit; bit >>= 1) j -= bit;
    j += bit;
    if (i < j) swap(a[i], a[j]);
  }
  for (int len = 2; len <= n; len <<= 1) {
    double ang = 2 * PI / len * (invert ? -1 : 1);
    base wlen(cos(ang), sin(ang));
    for (int i = 0; i < n; i += len) {
      w[0] = 1;
      for (int j = 0; j < len / 2; ++j) {
        if (j & 1) w[j] = w[j - 1] * wlen;
        if (j && !(j & 1)) w[j] = w[j >> 1] * w[j >> 1];
        base u = a[i + j], v = (a[i + j + len / 2] * w[j]);
        a[i + j] = (u + v);
        a[i + j + len / 2] = (u - v);
      }
    }
  }
  if (invert)
    for (int i = 0; i < n; ++i) a[i] /= n;
}
void multiply(vector<double>& a, vector<double> b) {
  vector<base> fa(a.begin(), a.end()), fb(b.begin(), b.end());
  int fsz = a.size() + b.size() - 1;
  int n = 1;
  while (n < fsz) n <<= 1;
  fa.resize(n), fb.resize(n);
  fft(fa, false), fft(fb, false);
  for (size_t i = 0; i < n; ++i) fa[i] *= fb[i];
  fft(fa, true);
  a.resize(fsz);
  for (int i = 0; i < fsz; i++) a[i] = fa[i].real();
}
vector<double> mul(vector<double>& a, vector<double>& b) {
  vector<double> c = a;
  multiply(c, b);
  return c;
}
vector<double> polymul(vector<vector<double> >& vec) {
  if (vec.size() == 1) return vec[0];
  int half = ((int)vec.size()) >> 1;
  vector<vector<double> > lft(vec.begin(), vec.begin() + half);
  vector<vector<double> > rgt(vec.begin() + half, vec.end());
  vector<double> vl = polymul(lft), vr = polymul(rgt);
  return mul(vl, vr);
}
vector<vector<double> > vec;
int main() {
  int n, m, x;
  scanf("%d", &(n)), scanf("%d", &(m));
  if (m == 1) {
    printf("1.0000000000\n");
    return 0;
  }
  int s = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &(x));
    x--;
    s += x;
    vector<double> v;
    for (int j = 0; j < m; j++) {
      v.push_back(j == x ? 0 : (1. / (m - 1)));
    }
    vec.push_back(v);
  }
  vector<double> v = polymul(vec);
  double ans = 0;
  for (int i = 0; i < s; i++) ans += v[i];
  printf("%.10lf\n", 1 + ans * (m - 1));
}