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: Runtime Error)

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=200010,M=100010;
int n,m,q;
int v[N],w[N],d[N],fa[N];
long long ans=0ll;
long long f[100][M];
inline void dfs1(int x)
{
    d[x]=d[x/2]+1;
    for(int k=w[x];k<=1e5;k++)
        f[x][k]=max(f[x][k],f[x/2][k-w[x]]+v[x]);
    if(d[x]==m) return;
    dfs1(x<<1);
    dfs1(x<<1|1);
    return;
}
inline void dfs2(int x,int y,long long z)
{
    if(d[x]==m)
    {
        ans=max(ans,z+f[x][y]);
        return;
    }
    if(y>=w[x]) dfs2(x/2,y-w[x],z+v[x]);
    dfs2(x/2,y,z);
    return;
}
int main()
{
    scanf("%d",&n);
    m=log2(n)/2;
    if(m==0) m=1;
    for(int k=1;k<=n;k++)
        scanf("%d%d",&v[k],&w[k]);
    dfs1(1);
    scanf("%d",&q);
    while(q--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        ans=0ll;
        if(d[x]) printf("%lld\n",f[x][y]);
        else dfs2(x,y,0ll),printf("%lld\n",ans);
    }
    return 0;
}