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;
ll cnt[100005],upper[100005],a[100005];
int n;
bool f=1;


int main() {
	cin>>n;
	for(int i=0;i<=n;i++) {
		cin>>a[i];
	}
	upper[n]=0;
	for(int i=n-1;i>=0;i--) {
		upper[i]=upper[i+1]+a[i+1];
	}
	cnt[0]=1-a[0];
	ll ans=1;
	if(cnt[0]>upper[0] || a[0]>1) {
		f=0;
	}
	for(int i=1;i<=n;i++) {
		if(2*cnt[i-1]<a[i]) {
			f=0;
			break;
		}
		cnt[i]=min(upper[i],2*cnt[i-1]-a[i]);
		ans+=cnt[i]+a[i];
	}
	if(!f) {
		cout<<-1;
		return 0;
	}
	cout<<ans;
	return 0;
}