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 Rust (1.42.0) (judge verdict: Accepted)

fn main() {
    proconio::input! {
        n: usize,
        vw: [(u32,usize);n],
        q: usize,
        vl: [(usize,usize);q],
    }
    let k=n.min(1<<9);
    let m=vl.iter().max_by_key(|vl|vl.1).unwrap().1 as usize;
    let mut dp=vec![vec![0;m+1];k+1];
    for (i,&(v,w)) in vw[0..k].iter().enumerate() {
        let i=i+1;
        let p=i/2;
        for j in 0..=m {
            let mut x=dp[p][j];
            if j>=w {
                x=x.max(dp[p][j-w]+v);
            }
            dp[i][j]=x;
        }
    }
    let mut u=Vec::with_capacity(9);
    let mut s=Vec::with_capacity(1<<9);
    for (v,l) in vl {
        u.clear();
        let mut v=v;
        while v>k {
            u.push(v-1);
            v/=2;
        }
        let mut r=dp[v][l];
        s.clear();
        s.push((0,0));
        for b in 1usize..1<<u.len() {
            let i=b.trailing_zeros()as usize;
            let (v2,w2)=s[b&b-1];
            let vw=vw[u[i]];
            let vw=(vw.0+v2,vw.1+w2);
            if vw.1<=l {
                r=r.max(dp[v][l-vw.1]+vw.0);
            }
            s.push(vw);
        }
        println!("{}",r);
    }
}