#include "biscuits.h"
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i=a; i<(b); ++i)
#define all(a) begin(a), end(a)
#define sz(x) (int)(x).size()
#define fi first
#define se second
#define pb push_back
template<typename T> using v = vector<T>;
template<typename T> using vv = v<vector<T>>;
typedef long long ll;
typedef pair<int ,int > pii;
typedef v<int> vi;
//Represent y as a subset of indices
//yi = {0,x}
//a0 >= y0
//a1 + a0/2 >= y1 + y0/2
//a2 + a1/2 + a0 >= y2 + y1/2 + y0/4
//...
//mutiply the ith eqn by 2^i, we get
//a0 >= y0
//2a1 + a0 >= 2y1 + y0
//4a2 + 2a1 + a0 >= 4y2 + 2y1 + y0
//2^k*ak + .... + a0 >= 2^k*yk + ... y0
//Let S = set of possible values of y0+2y1+4y2 + ... 2^i * yi
//after processing a0 to ai
//we need to do two things
//1) S = S U 2^i * x
//2) truncate S by p(i) = sum_j<=i 2^j * aj
//Obs: gap between segments <= 2^i, implying the number of segments is either 1 or 2
const ll INF = 2e18;
ll count_tastiness(ll x, vector<ll> a) {
ll k = sz(a);
vector<ll>p(60,0);
for(int i=0;i<k;i++)p[i] = a[i]* (1LL<<i);
for(int i=1;i<60;i++)p[i] += p[i-1];
//for(ll x:p)cout<<x<<" ";
//cout<<'\n';
set<ll,greater<ll>>s;
s.insert(0);
for(int i=0;i<60;i++){
if(x >= INF/(1LL<i)){}
else{
vector<ll>t;
for(ll v:s){
ll val = v + (1LL<<i)*x;
if(val <= p[i])t.push_back(val);
}
for(ll v:t)s.insert(v);
}
while(!s.empty() && *s.begin() > p[i])s.erase(s.begin());
//cout<<"At i\n";
//for(ll x:s)cout<<x<<" ";cout<<'\n';
}
return sz(s);
}
