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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

const int MOD=1000000007;
//~ const int MOD=998244353;
const int INF=0x3f3f3f3f;
const ll INFL=0x3f3f3f3f3f3f3f3f;

int a[200000];
ll L[200000];

int main(){
	int n;cin>>n;
	rep(i,n+1)scanf("%d",&a[i]);
	if(n==0){
		if(a[0]==1)puts("1");
		else puts("-1");
		return 0;
	}
	L[0]=1;
	rep(i,n+1){
		L[i+1]=max(0LL,min(INFL,(L[i]-a[i])*2));
	}
	ll cnt=0;
	ll ans=0;
	for(int i=n;i>=0;i--){
		if((cnt+1)/2+a[i]>L[i]){
			puts("-1");return 0;
		}
		cnt=min(L[i],cnt+a[i]);
		ans+=cnt;
	}
	cout<<ans<<endl;
}