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<stdio.h>
#include<stdlib.h>

int stack_w[18],stack_v[18],sp=0;
int dp[19][100001];

int max(int a, int b){
  if(a>=b)return a;
  else return b;
}

void stack_init(int *V,int *W,int num){	
  sp=0;
  while(num>0){
    stack_v[sp]=V[num],stack_w[sp]=W[num];
    sp++;
  	num=num/2;
  }
}

int rec(int i, int j){
  if(dp[i][j]!=-1){
  	return dp[i][j];
  }
	int res;
  if(i==sp){
  	res=0;
  }
  else if(j<stack_w[i]){
  	res=rec(i+1,j);
  }
  else{
  	res=max(rec(i+1,j),rec(i+1,j-stack_w[i])+stack_v[i]);
  }
  return dp[i][j]=res;
}

int main(){

	int N,i,j,k;
  scanf("%d",&N);
  int *V=(int*)malloc(sizeof(int)*(N+1));
  int *W=(int*)malloc(sizeof(int)*(N+1));
  for(i=1;i<N+1;i++)scanf("%d %d",&V[i],&W[i]);
  int Q;
  scanf("%d",&Q);
  int *v=(int*)malloc(sizeof(int)*Q);
  int *L=(int*)malloc(sizeof(int)*Q); 
  for(i=0;i<Q;i++)scanf("%d %d",&v[i],&L[i]);
  for(i=0;i<19;i++){
    for(j=0;j<100001;j++)dp[i][j]=-1;
  }
  for(k=0;k<Q;k++){
  	stack_init(V,W,v[i]);
    printf("%d\n",rec(0,L[i]));
  }
  return 0;
}