Problem:
Score : 700 points

Problem Statement

We have a rooted binary tree with N vertices, where the vertices are numbered 1 to N.
Vertex 1 is the root, and the parent of Vertex i (i \geq 2) is Vertex \left[ \frac{i}{2} \right].

Each vertex has one item in it. The item in Vertex i has a value of V_i and a weight of W_i.
Now, process the following query Q times:

Given are a vertex v of the tree and a positive integer L.
Let us choose some (possibly none) of the items in v and the ancestors of v so that their total weight is at most L.
Find the maximum possible total value of the chosen items.

Here, Vertex u is said to be an ancestor of Vertex v when u is an indirect parent of v, that is, there exists a sequence of vertices w_1,w_2,\ldots,w_k (k\geq 2) where w_1=v, w_k=u, and w_{i+1} is the parent of w_i for each i.

Constraints

All values in input are integers.

1 \leq N < 2^{18}

1 \leq Q \leq 10^5

1 \leq V_i \leq 10^5

1 \leq W_i \leq 10^5

For the values v and L given in each query, 1 \leq v \leq N and 1 \leq L \leq 10^5.

Input

Let v_i and L_i be the values v and L given in the i-th query.
Then, Input is given from Standard Input in the following format:

N
V_1 W_1
:
V_N W_N
Q
v_1 L_1
:
v_Q L_Q

Output

For each integer i from 1 through Q,
the i-th line should contain the response to the i-th query.

Sample Input 1

3
1 2
2 3
3 4
3
1 1
2 5
3 5

Sample Output 1

0
3
3

In the first query, we are given only one choice: the item with (V, W)=(1,2). Since L = 1, we cannot actually choose it, so our response should be 0.

In the second query, we are given two choices: the items with (V, W)=(1,2) and (V, W)=(2,3). Since L = 5, we can choose both of them, so our response should be 3.

Sample Input 2

15
123 119
129 120
132 112
126 109
118 103
115 109
102 100
130 120
105 105
132 115
104 102
107 107
127 116
121 104
121 115
8
8 234
9 244
10 226
11 227
12 240
13 237
14 206
15 227

Sample Output 2

256
255
250
247
255
259
223
253

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

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef pair<ll,ll> pa;

#define fst first
#define snd second
#define pb push_back

const int maxn=500005,maxm=512;
const double pi=acos(-1.0);
//const ll mod=1000000007;
const ll mod=998244353;
const double eps=1e-8;

int dig(int xx)
{
    for(int i=0;;i++)
    {
        if(xx>>i&1) return i;
    }
}


struct dat
{
    int sta;
    ll v,w;
    dat(ll v=0,ll w=0):sta(sta),v(v),w(w){}
};

int n,q;
int v[maxn]={0},w[maxn]={0};

ll ve[maxm][maxm][2]={0};

vector<pa> ans1[maxm];

vector<pa> tmp;

ll cl2(int vv,int ww)
{
    int lb=0,ub=ans1[vv].size();
    while(ub-lb>1)
    {
        int mid=(ub+lb)/2;
        if(ans1[vv][mid].fst<=ww) lb=mid;
        else ub=mid;
    }
    return ans1[vv][lb].snd;
}


void solve(int vv,int L)
{
    int i,j;
    if(vv<512)
    {
        for(j=0;;j++) {if((vv<<j)>=maxm/2&&(vv<<j)<maxm) break;}
        ll ans=0;
        for(i=1;i<(1<<(j+1));i++)
        {
            if(ve[vv<<j][i][1]<=L) ans=max(ans,ve[vv<<j][i][0]);
        }
        printf("%lld\n",ans);return;
    }
    tmp.clear();tmp.pb(pa(0,0));if(w[vv]<=L) tmp.pb(pa(w[vv],v[vv]));
    while(1)
    {
        vv/=2;
        if(vv>=maxm/2&&vv<maxm) break;
        int nn=tmp.size();
        for(i=0;i<nn;i++)
        {
            if(tmp[i].fst+w[vv]<=L) {tmp.pb(pa(tmp[i].fst+w[vv],tmp[i].snd+v[vv]));}
        }
    }
    ll ans=0;
    for(i=0;i<tmp.size();i++)
    {
        ll www=tmp[i].fst,vvv=tmp[i].snd;
        ans=max(ans,cl2(vv,L-www)+vvv);
    }
    printf("%lld\n",ans);
}

bool cmp(pa a,pa b)
{
    if(a.fst!=b.fst) return a.fst<b.fst;
    return a.snd<b.snd;
}

int tp[maxn];

int main()
{
	int i,j;
	scanf("%d",&n);
	for(i=1;i<=n;i++)
    {
        scanf("%d %d",v+i,w+i);
    }
    n=max(n,maxm-1);
    for(i=maxm/2;i<maxm;i++)
    {
        for(j=1;j<maxm;j++)
        {
            int jj=j&(-j);jj=j-jj;
            ve[i][j][0]=ve[i][jj][0]+v[i>>(8-dig(j))];
            ve[i][j][1]=ve[i][jj][1]+w[i>>(8-dig(j))];
        }
    }
    for(i=maxm/2;i<maxm;i++)
    {
        for(j=0;j<maxm;j++) ans1[i].pb(pa(ve[i][j][1],ve[i][j][0]));
        sort(ans1[i].begin(),ans1[i].end(),cmp);
    }

    scanf("%d",&q);
    for(i=0;i<q;i++)
    {
        int vv,L;
        scanf("%d %d",&vv,&L);
        solve(vv,L);
    }
    return 0;
}