Problem:
Score : 600 points

Problem Statement

Given is an integer sequence of length N+1: A_0, A_1, A_2, \ldots, A_N. Is there a binary tree of depth N such that, for each d = 0, 1, \ldots, N, there are exactly A_d leaves at depth d? If such a tree exists, print the maximum possible number of vertices in such a tree; otherwise, print -1.

Notes

A binary tree is a rooted tree such that each vertex has at most two direct children.

A leaf in a binary tree is a vertex with zero children.

The depth of a vertex v in a binary tree is the distance from the tree's root to v. (The root has the depth of 0.)

The depth of a binary tree is the maximum depth of a vertex in the tree.

Constraints

0 \leq N \leq 10^5

0 \leq A_i \leq 10^{8} (0 \leq i \leq N)

A_N \geq 1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_0 A_1 A_2 \cdots A_N

Output

Print the answer as an integer.

Sample Input 1

3
0 1 1 2

Sample Output 1

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print 7.

Sample Input 2

4
0 0 1 0 2

Sample Output 2

10

Sample Input 3

2
0 3 1

Sample Output 3

-1

Sample Input 4

1
1 1

Sample Output 4

-1

Sample Input 5

10
0 0 1 1 2 3 5 8 13 21 34

Sample Output 5

264

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;


#define rep(i,n) for(ll (i)=0; (i)<(ll)(n); (i)++)
#define frep(i,m,n) for(ll (i)=(m); (i)<=(ll)(n); (i)++)
#define rrep(i,n) for(ll (i)=(n)-1; (i)>-1; (i)--)
#define frrep(i,m,n) for(ll (i)=(n); (i)>(ll)(m); (i)--)
#define ALL(x) (x).begin(), (x).end()

const ll INF = 100100100100100100;
const ll MOD = 1000000007;

// get abs
ll my_abs(ll a);
// a^n
ll a_n(ll a, ll n);
// get gcd
ll my_gcd(ll a, ll b);
// a^(-1) % MOD 
ll inv(ll a);
// (a+b+c)%MOD
ll madd(ll a, ll b, ll c);
// (a-b)%MOD
ll msub(ll a, ll b);
// (a*b*c)%MOD
ll mtime(ll a, ll b, ll c);


int main() {
    ll n; cin >> n;
    // x[i]: 深さiで葉でない, sum[i]: a[i]までの累積和
    vector<ll> a(n+1), b(n+1, 0), sum(n+1, 0);
    rep(i, n+1) cin >> a[i];
    sum[0] = a[0];
    rep(i, n) {
        sum[i+1] = sum[i] + a[i+1];
    }
    ll ans = 0;

    bool ok = true;
    rep(i, n+1) {
        if(i == 0) {
            b[i] = 1-a[i];
            ans++;
        }
        else {
            ll ub = min(2*b[i-1]-a[i], sum[n]-sum[i]);
            if(b[i-1]-a[i] > ub) {
                ok = false;
                break;
            }
            else {
                b[i] = ub;
                ans += (a[i] + b[i]);
            }
        }
    }

    if(ok) {
        cout << ans << endl;
    }
    else {
        cout << -1 << endl;
    }

    return 0;
}

ll my_abs(ll a) {
    if(a >= 0) return a;
    else return -1 *a;
}

ll a_n(ll a, ll n) { 
    if(n == 0) return 1;
    
    ll ret = a, count = 1;
    while(count * 2 < n) {
        ret *= ret;
        count *= 2;
    }
    if(count == n) return ret;
    else return (ret * a_n(a, n-count));
}

ll my_gcd(ll a, ll b) {
    if(b == 0) return a;
    return my_gcd(b, a%b);
}

ll inv(ll a) {
    return a_n(a, MOD-2);
}

ll madd(ll a, ll b, ll c) {
    ll ret = (a+b) % MOD;
    return (ret+c) % MOD;
}

ll msub(ll a, ll b) {
    if(a < b) return (a-b+MOD) % MOD;
    else return (a-b) % MOD;
}

ll mtime(ll a, ll b, ll c) {
    ll ret = (a*b) % MOD;
    return (ret*c) % MOD;
}