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 Java (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.IOException;
import java.io.InputStream;
import java.util.HashMap;
import java.util.Map.Entry;
import java.util.NoSuchElementException;

public class Main {
    static int N,Q;
    static int[] V,W,U,L;
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        N = (int)sc.nextLong();
        V = new int[N];
        W = new int[N];
        for(int i=0; i<N; i++) {
            V[i] = (int)sc.nextLong();
            W[i] = (int)sc.nextLong();
        }
        Q = (int)sc.nextLong();
        U = new int[Q];
        L = new int[Q];
        for(int i=0; i<Q; i++) {
            U[i] = (int)sc.nextLong();
            L[i] = (int)sc.nextLong();
        }

        HashMap<Integer, Integer> dp,dp2;
        for(int i=0; i<Q; i++) {
            dp = new HashMap<>();
            dp.put(0, 0);
            int u = U[i];
            int l = L[i];
            while (u > 0) {
                int v = V[u-1];
                int w = W[u-1];
                dp2 = new HashMap<>();
                for (Entry<Integer, Integer> e: dp.entrySet()) {
                    int w0 = e.getKey();
                    int v0 = e.getValue();
                    if(dp2.containsKey(w0)) {
                        dp2.put(w0, Math.max(dp2.get(w0), v0));
                    } else {
                        dp2.put(w0, v0);
                    }
                    if (w0+w > l) continue;
                    if(dp2.containsKey(w0+w)) {
                        dp2.put(w0+w, Math.max(dp2.get(w0+w), v0+v));
                    } else {
                        dp2.put(w0+w, v0+v);
                    }
                }
                dp = dp2;
                u /= 2;
            }
            int ans = 0;
            for (Entry<Integer, Integer> e: dp.entrySet()) {
                ans = Math.max(ans, e.getValue());
            }
            System.out.println(ans);
        }

    }

    //https://qiita.com/p_shiki37/items/65c18f88f4d24b2c528b
    static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;
        private boolean hasNextByte() {
            if (ptr < buflen) {
                return true;
            }else{
                ptr = 0;
                try {
                    buflen = in.read(buffer);
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (buflen <= 0) {
                    return false;
                }
            }
            return true;
        }
        private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
        private boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
        private void skipUnprintable() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;}
        public boolean hasNext() { skipUnprintable(); return hasNextByte();}
        public String next() {
            if (!hasNext()) throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while(isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }
        public long nextLong() {
            if (!hasNext()) throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) {
                throw new NumberFormatException();
            }
            while(true){
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                }else if(b == -1 || !isPrintableChar(b)){
                    return minus ? -n : n;
                }else{
                    throw new NumberFormatException();
                }
                b = readByte();
            }
        }
    }
}