Problem: You're playing a game called Osu! Here's a simplified version of it. There are n clicks in a game. For each click there are two outcomes: correct or bad. Let us denote correct as "O", bad as "X", then the whole play can be encoded as a sequence of n characters "O" and "X".

Using the play sequence you can calculate the score for the play as follows: for every maximal consecutive "O"s block, add the square of its length (the number of characters "O") to the score. For example, if your play can be encoded as "OOXOOOXXOO", then there's three maximal consecutive "O"s block "OO", "OOO", "OO", so your score will be 22 + 32 + 22 = 17. If there are no correct clicks in a play then the score for the play equals to 0.

You know that the probability to click the i-th (1 ≤ i ≤ n) click correctly is pi. In other words, the i-th character in the play sequence has pi probability to be "O", 1 - pi to be "X". You task is to calculate the expected score for your play.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of clicks. The second line contains n space-separated real numbers p1, p2, ..., pn (0 ≤ pi ≤ 1).

There will be at most six digits after the decimal point in the given pi.

Output

Print a single real number — the expected score for your play. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Examples

Input

3
0.5 0.5 0.5


Output

2.750000000000000


Input

4
0.7 0.2 0.1 0.9


Output

2.489200000000000


Input

5
1 1 1 1 1


Output

25.000000000000000

Note

For the first example. There are 8 possible outcomes. Each has a probability of 0.125.

  * "OOO"  →  32 = 9; 
  * "OOX"  →  22 = 4; 
  * "OXO"  →  12 + 12 = 2; 
  * "OXX"  →  12 = 1; 
  * "XOO"  →  22 = 4; 
  * "XOX"  →  12 = 1; 
  * "XXO"  →  12 = 1; 
  * "XXX"  →  0. 



So the expected score is <image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const double eps = 1e-12;
const int inf = 2000000000;
const long long int infLL = (long long int)2e18;
long long int MOD = 998244353;
int MOD1 = 1000000007;
int MOD2 = 1000000009;
inline bool checkBit(long long int n, long long int i) {
  return n & (1LL << i);
}
inline long long int setBit(long long int n, long long int i) {
  return n | (1LL << i);
  ;
}
inline long long int resetBit(long long int n, long long int i) {
  return n & (~(1LL << i));
}
int dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline bool isLeapYear(long long int year) {
  return (year % 400 == 0) || (year % 4 == 0 && year % 100 != 0);
}
inline void normal(long long int &a) {
  a %= MOD;
  (a < 0) && (a += MOD);
}
inline long long int modMul(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a * b) % MOD;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  return (a + b) % MOD;
}
inline long long int modSub(long long int a, long long int b) {
  a %= MOD, b %= MOD;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1LL;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1LL;
  }
  return r;
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modPow(b, MOD - 2));
}
bool comp(const pair<long long int, pair<long long int, long long int> > &p1,
          const pair<long long int, pair<long long int, long long int> > &p2) {
  return p1.first > p2.first;
}
bool comp1(const pair<long long int, long long int> &p1,
           const pair<long long int, long long int> &p2) {
  if (p1.first == p2.first) {
    return p1.second > p2.second;
  }
  return p1.first < p2.first;
}
long long int converter(string a) {
  long long int i, mul = 1, r, t, ans = 0LL;
  if (a.length() == 0) return 0;
  for (i = a.length() - 1; i >= 0; i--) {
    t = a[i] - '0';
    r = t % 10;
    ans += (mul * r);
    mul = mul * 10;
  }
  return ans;
}
int msb(unsigned x) {
  union {
    double a;
    int b[2];
  };
  a = x;
  return (b[1] >> 20) - 1023;
}
const int MAX = 100005;
double a[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cout.unsetf(ios::floatfield);
  cout.precision(14);
  cout.setf(ios::fixed, ios::floatfield);
  ;
  int n, i;
  cin >> n;
  double allsum = 0.0, res = 0.0;
  for (i = 1; i <= n; ++i) cin >> a[i];
  for (i = 1; i <= n; ++i) {
    res += a[i];
    allsum = (allsum + a[i - 1]) * a[i];
    res += (allsum * 2.0);
  }
  cout << res << '\n';
  return 0;
}