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 JavaScript (Node.js 12.16.1) (judge verdict: Time Limit Exceeded)

"use strict";
var input=require("fs").readFileSync("/dev/stdin","utf8");
var cin=input.split(/ |\n/),cid=0;
function next(){return +cin[cid++];}
function nextstr(){return cin[cid++];}
function nextbig(){return BigInt(cin[cid++]);}
function nexts(n,a){return a?cin.slice(cid,cid+=n):cin.slice(cid,cid+=n).map(a=>+a);}
function nextm(h,w,a){var r=[],i=0;if(a)for(;i<h;i++)r.push(cin.slice(cid,cid+=w));else for(;i<h;i++)r.push(cin.slice(cid,cid+=w).map(a=>+a));return r;}
function xArray(v){var a=arguments,l=a.length,r="Array(a["+--l+"]).fill().map(x=>{return "+v+";})";while(--l)r="Array(a["+l+"]).fill().map(x=>"+r+")";return eval(r);}

var myOut = main();
if(myOut !== undefined)console.log(String(myOut));

function main(){
  var n = next();
  var v = nextm(n,2);
  var w = Array(n)
  for(var i = 0; i < n; i++){
    var d = 31-Math.clz32(i+1);
    if(d === 0 || d === 9){
      w[i] = [v[i]];
    }else{
      var t = i-1>>1;
      w[i] = w[t].concat();
      w[i].push(v[i]);
      for(var j = 0; j < w[t].length; j++){
        w[i].push([w[t][j][0]+v[i][0],w[t][j][1]+v[i][1]]);
      }
      w[i].sort((a,b)=>a[1]-b[1]);
    }
  }
  var q = next();
  var ans = Array(q);
  for(var i = 0; i < q; i++){
    var x = next(), s = next();
    var d = 31-Math.clz32(x);
    var z = 0;
    if(d < 9){
      for(var j = 0; j < w[x-1].length && w[x-1][j][1] <= s; j++){
        if(z < w[x-1][j][0])z = w[x-1][j][0];
      }
      ans[i] = z;
    }else{
      var r = w[x-1].length, y = x;
      while(d < 9)d--, y >>= 1;
      y--;
      for(var j = 0; j < w[y].length; j++){
        while(r >= 0 && w[x-1][r][1]+w[y][j][1] > s)r--;
        if(r < 0)break;
        if(z < w[x-1][r][0]+w[y][j][0])z = w[x-1][r][0]+w[y][j][0];
      }
      ans[i] = z;
    }
  }
  return ans.join("\n");
}