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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define FOR(i,l,r) for(long long i=(l);i<(r);++i)
#define REP(i,n) FOR(i,0,n)
#define REPS(i,n) FOR(i,1,n+1)
#define RFOR(i,l,r) for(long long i=(l);i>=(r);--i)
#define RREP(i,n) RFOR(i,n-1,0)
#define RREPS(i,n) RFOR(i,n,1)
#define int long long
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b; return true; }return false; }
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b; return true; }return false; }
const int INF = 1e18;

int dep(int n){
    int ret = 0;
    while(n >= 2){
        n /= 2;
        ret++;
    }
    return ret;
}
int func(int n,int j){
    while(j--) n /= 2;
    return n;
}

signed main(){
    //input
    int n; cin >> n;
    int v[n+1],w[n+1];
    REPS(i,n) cin >> v[i] >> w[i];
    int q; cin >> q;
    int M_L = 0;
    int t[q], l[q];
    REP(i,q){
        cin >> t[i] >> l[i];
        chmax(M_L,l[i]);
    }
    //半分全列挙
    int V = min((1ll << 9)-1, n);
    vector<vector<int>> dp(V+1,vector<int>(M_L+1,0));
    REPS(i,V) REP(j,M_L+1){
        chmax(dp[i][j], dp[i/2][j]);
        if(j - w[i] > -1) chmax(dp[i][j], dp[i/2][j-w[i]] + v[i]);
    }
    REP(i,q){
        if(t[i] <= V) cout << dp[t[i]][l[i]] << '\n';
        else{
            int z = dep(t[i]) - 8;
            int o = func(t[i],z);
            int ans = 0;
            for(int b = 0; b < (1ll << z); b++){
                int sumw = 0;
                int sumv = 0;
                REP(j,z) if((b >> j) & 1){
                    sumw += w[func(t[i],j)];
                    sumv += v[func(t[i],j)];
                }
                if(l[i]-sumw > -1) chmax(ans, dp[o][l[i]-sumw] + sumv);
            }
            cout << ans << '\n';
        }
    }
}