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 INTINF 1999999999
#define REP(i,n) for(int i=0;i<n;i++)
#define REP1(i,n) for(ll i=1;i<=n;i++)
#define MODA 1000000007 

int main() {
	int tmp;
	ll ans = 0;
	int n;
	cin >> n;
	ll a[n+1], sum = 0;
	REP(i, n+1){
		cin >> a[i];
		sum += a[i];
	}
	bool ch = true;
	ll now = 1;
	REP(i, n+1){
		if(now < a[i]) ch = false;
		ans += now;
		sum -= a[i];
		now -= a[i];
		now = min(sum, now * 2);
	}
	if(ch)cout << ans << endl;
	else cout << -1 << endl;
}
