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();
  const aa = 9;
  var mn = Math.min(n,1<<aa);
  const mx = 1e5+1;
  var dp = Array(mn);
  var V = nextm(n,2);
  var v = V[0][0], w = V[0][1];
  dp[0] = Array(mx).fill(0,0,w).fill(v,w);
  for(var i = 1; i < mn; i++){
    var p = i-1>>1;
    v = V[i][0], w = V[i][1];
    dp[i] = dp[p].concat();
    for(var j = mx-1; w <= j; j--)if(dp[i][j] < dp[i][j-w]+v)dp[i][j] = dp[i][j-w]+v;
  }
  var q = next();
  var ans = Array(q);
  for(var i = 0; i < q; i++){
    var [u,l] = nexts(2);
    var d = 31-Math.clz32(u--);
    var s = [];
    while(aa < d){
      v = V[u][0], w = V[u][1];
      if(l < w)continue;
      var k = s.length;
      s.push(v, w);
      for(var j = 0; j < k; j+=2){
        var t = s[j+1]+w;
        if(t <= l){
          s.push(s[j]+v, t);
        }
      }
      u = u-1>>1, d--;
    }
    ans[i] = d < 9 ? dp[u][l] : dp[u-1>>1][l];
    if(aa === d && l >= V[u][1]){
      v = V[u][0], w = V[u][1];
      u = u-1>>1
      if(ans[i] < dp[u][l-w]+v)ans[i] = dp[u][l-w]+v;
      for(var j = 0; j < s.length; j+=2){
        if(ans[i] < dp[u][l-s[j+1]]+s[j])ans[i] = dp[u][l-s[j+1]]+s[j];
        var t = s[j+1]+w;
        if(t <= l){
          if(ans[i] < dp[u][l-t]+s[j]+v)ans[i] = dp[u][l-t]+s[j]+v;
        }
      }
    }
  }
  return ans.join("\n");
}