Problem:
Score : 600 points

Problem Statement

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.

Notes

A binary tree is a rooted tree such that each vertex has at most two direct children.

A leaf in a binary tree is a vertex with zero children.

The depth of a vertex v in a binary tree is the distance from the tree's root to v. (The root has the depth of 0.)

The depth of a binary tree is the maximum depth of a vertex in the tree.

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.

Sample Input 1

3
0 1 1 2

Sample Output 1

7

Below is the tree with the maximum possible number of vertices. It has seven vertices, so we should print 7.

Sample Input 2

4
0 0 1 0 2

Sample Output 2

10

Sample Input 3

2
0 3 1

Sample Output 3

-1

Sample Input 4

1
1 1

Sample Output 4

-1

Sample Input 5

10
0 0 1 1 2 3 5 8 13 21 34

Sample Output 5

264

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<stdio.h>
#include<vector>
#include<queue>
#include<math.h>
#include<deque>
using namespace std;
#define long long ll
#define rep(s,i,n) for(int i=s;i<n;i++)
#define c(n) cout<<n<<endl;
#define ic(n) int n;cin>>n;
#define sc(s) string s;cin>>s;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c,a,b) *min_element(c+a,c+b)
#define maxi(c,a,b) *max_element(c+a,c+b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int,int>
#define upp(a,n,x) upper_bound(a,a+n,x)-a;
#define low(a,n,x) lower_bound(a,a+n,x)-a;
#define UF UnionFind 
//printf("%.12Lf\n",);
int keta(int x) {
	rep(0, i, 30) {
		if (x < 10) {
			return i + 1;
		}
		x = x / 10;
	}
}
int gcd(int x, int y) {
	if (x == 0 || y == 0)return x + y;
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return bb;
		}
		bb = bb % aa;
		if (bb == 0) {
			return aa;
		}
	}
}
int lcm(int x, int y) {
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return x / bb * y;
		}
		bb = bb % aa;
		if (bb == 0) {
			return x / aa * y;
		}
	}
}
bool p(int x) {
	if (x == 1)return false;
	rep(2, i, sqrt(x) + 1) {
		if (x % i == 0 && x != i) {
			return false;
		}
	}
	return true;
}
int max(int a, int b) {
	if (a >= b)return a;
	else return b;
}
string maxst(string s, string t) {
	int n = s.size();
	int m = t.size();
	if (n > m)return s;
	else if (n < m)return t;
	else {
		rep(0, i, n) {
			if (s[i] > t[i])return s;
			if (s[i] < t[i])return t;
		}
		return s;
	}
}
int min(int a, int b) {
	if (a >= b)return b;
	else return a;
}
int n2[41];
int nis[41];
int nia[41];
int mody[41];
int nn;
int com(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int bunsi = 1, bunbo = 1;
	rep(0, i, y)bunsi = (bunsi * (n - i)) % mod;
	rep(0, i, y)bunbo = (bunbo * (i + 1)) % mod;
	mody[0] = bunbo;
	rep(1, i, 41) {
		bunbo = (bunbo * bunbo) % mod;
		mody[i] = bunbo;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = mod - 2;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 41) {
		if (nis[i] == 1) {
			bunsi = (bunsi * mody[i]) % mod;
		}
	}
	return bunsi;
}
int gyakugen(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	mody[0] = y;
	rep(1, i, 41) {
		y = (y * y) % mod;
		mody[i] = y;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = mod - 2;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 41) {
		if (nis[i] == 1) {
			n = (n * mody[i]) % mod;
		}
	}
	return n;
}
int yakuwa(int n) {
	int sum = 0;
	rep(1, i, sqrt(n + 1)) {
		if (n % i == 0)sum += i + n / i;
		if (i * i == n)sum -= i;
	}
	return sum;
}
int poow(int y, int n) {
	if (n == 0)return 1;
	n -= 1;
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int yy = y;
	mody[0] = yy;
	rep(1, i, 41) {
		yy = (yy * yy) % mod;
		mody[i] = yy;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = n;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn >= n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	rep(0, i, 41) {
		if (nis[i] == 1) {
			y = (y * mody[i]) % mod;
		}
	}
	return y;
}
int minpow(int x, int y) {
	int sum = 1;
	rep(0, i, y)sum *= x;
	return sum;
}
int ketawa(int x, int sinsuu) {
	int sum = 0;
	rep(0, i, 100)sum += (x % poow(sinsuu, i + 1)) / (poow(sinsuu, i));
	return sum;
}
int sankaku(int a) {
	return a * (a + 1) / 2;
}
int sames(int a[1111111], int n) {
	int ans = 0;
	rep(0, i, n) {
		if (a[i] == a[i + 1]) {
			int j = i;
			while (a[j + 1] == a[i] && j <= n - 2)j++;
			ans += sankaku(j - i);
			i = j;
		}
	}
	return ans;
}
/*using Graph=vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph &G, int v, int p, int d) {
	depth[v] = d;
	oya[v]=p;
	for (auto nv : G[v]) {
	   if (nv == p) continue; // nv が親 p だったらダメ
	   dfs(G, nv, v, d+1); // d を 1 増やして子ノードへ
	}
}*/
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
	memo[h][w] = 'x';

	// 八方向を探索
	for (int dh = -1; dh <= 1; ++dh) {
		for (int dw = -1; dw <= 1; ++dw) {
			if(abs(0-dh)+abs(0-dw)==2)continue;
			int nh = h + dh, nw = w + dw;

			// 場外アウトしたり、0 だったりはスルー
			if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
			if (memo[nh][nw] == 'x') continue;

			// 再帰的に探索
			dfs(nh, nw);
		}
	}
}*/
int XOR(int a, int b) {
	if (a == 0 || b == 0) {
		return a + b;
	}
	int ni = 1;
	rep(0, i, 41) {
		n2[i] = ni;
		ni *= 2;
	}
	rep(0, i, 41)nis[i] = 0;
	for (int i = 40;i >= 0;i -= 1) {
		if (a >= n2[i]) {
			nis[i]++;
			a -= n2[i];
		}
		if (b >= n2[i]) {
			nis[i]++;
			b -= n2[i];
		}
	}
	int sum = 0;
	rep(0, i, 41)sum += (nis[i] % 2 * n2[i]);
	return sum;
}
//int ma[1024577][21];
//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
	vector<int> siz;
	UnionFind(int N) :par(N), siz(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) {
			par[i] = i;
			siz[i] = 1;
		}
	}

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		if (siz[rx] > siz[ry]) {
			siz[rx] += siz[ry];
			par[y] = x;
		}
		else {
			siz[ry] += siz[rx];
			par[x] = y;
		}
	}
	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
	int size(int x) {
		return siz[root(x)];
	}
};
int a[514514];
int sum[514514];
signed main() {
	ic(n)
	sum[0] = 0;
	rep(0, i, n+1) {
		cin >> a[i];
		sum[i + 1] = sum[i] + a[i];
	}
	int ans = 1;
	int ty = 1;
	if (a[0] > 0) {
		c(-1)
		return 0;
	}
	rep(0, i, n + 1) {
		ans += ty;
		ty= ty - a[i];
		if (ty < 0) {
			c(-1)
			return 0;
		}
		ty = min(ty * 2, sum[n + 1] - sum[i]);
	}
	c(ans)
}