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;
#define fi first
#define se second
int n,m;
ll a[100001];
ll b[100001];
int main(){
	ios::sync_with_stdio(false);
	cin >> n;
	for(int i=0; i<=n ;i++){
		cin >> a[i];
		if(i!=0) b[i]=min((ll)1e18,b[i-1]*2-a[i]);
		else b[i]=1-a[i];
	}
	ll ans=0;
	ll cur=0;
	for(int i=n; i>=1 ;i--){
		cur+=a[i];
		ans+=cur;
		if(b[i-1]*2<cur) return cout << "-1\n",0;
		cur=min(cur,b[i-1]);
	}
	if(cur+a[0]!=1) cout << "-1\n";
	else cout << ans+1 << endl;
}