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: Accepted)

#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
using namespace std;
const int N=3e5+5,M=1e5+5;
int n,q,dep[N],p[N],w[N],dp[512][M];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d%d",&p[i],&w[i]);
    for(int i=1;i<=n&&i<512;i++)
        for(int j=100000;j>=0;j--)
        {
            if(j>=w[i]) dp[i][j]=max(dp[i][j],dp[i/2][j-w[i]]+p[i]);
            dp[i][j]=max(dp[i][j],dp[i/2][j]);
        }
    scanf("%d",&q);
    while(q--)
    {
        int v,L;scanf("%d%d",&v,&L);
        vector<int>vc;
        while(v>=512) vc.push_back(v),v>>=1;
        int ans=dp[v][L];
        for(int i=0;i<1<<vc.size();i++)
        {
            int wi=0,pi=0;
            for(int j=0;j<vc.size();j++)
                if(i>>j&1)
            {
                wi+=w[vc[j]];
                pi+=p[vc[j]];
                if(wi>L) break;
            }
            if(wi<=L)
                ans=max(ans,dp[v][L-wi]+pi);
        }
        printf("%d\n",ans);
    }
}