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.

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.

Examples

Input

3
0 1 1 2


Output

7


Input

4
0 0 1 0 2


Output

10


Input

2
0 3 1


Output

-1


Input

1
1 1


Output

-1


Input

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


Output

264
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main()
{

    int n;
    scanf("%d",&n);

    vector<ll> a(n+1);
    for (int i = 0; i < n+1; i++)
    {
        scanf("%lld",&a[i]);
    }
    ll sum = 0;
    for (int i = 0; i <= n; i++)
    {
        sum += a[i];
    }
    ll ans = 0, space = 1;
    for (ll i = 0; i <= n; i++, space *= 2, space = min(space, sum))
    {
        ans += min(sum, space);
        sum -= a[i];
        space -= a[i];
        if (space < 0)
        {
            printf("-1");
            return 0;
        }
    }

    printf("%lld\n",ans);
    return 0;
}

