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;
#define rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
int n, cnt, a[100005];
ll ac[100005], t = 1, ans;
int main(){
	scanf("%d", &n);
	rep(i,n+1) scanf("%d", a+i);
	for(int i=n; i>=0; i--) ac[i] = ac[i+1] + a[i];
	rep(i,n+1){
		ans += t;
		t -= a[i];
		if(t < 0) return puts("-1") * 0;
		t = min(t*2, ac[i+1]);
	}
	printf("%lld\n", ans);
}