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: Accepted)

#include<bits/stdc++.h>
#include<stdio.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
typedef unsigned long long int lu;
typedef vector< int > vi;
typedef vector < ll > vl;
typedef set <ll> sl;
typedef vector< vi > vvi;
typedef vector< vl > vvl;
typedef pair< int,int > ii;
typedef pair< ll,ll > il;
#define sz(a) ll((a).size())
#define test(t) ll t;cin>>t;while(t--)
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(__typeof((c).begin()) i = (c).begin(); i != (c).end(); i++)
#define present(c,x) ((c).find(x) != (c).end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define Vi vector
#define pr pair
#define mpr make_pair
#define fi first
#define se second
#define isnum(a) (a>=48 && a<=57)
#define isS(a) (a>='a' && a<='z')
#define isU(a) (a>='A' && a<='Z')
#define toS(a) (isU(a)?a+32:a)
#define toU(a) (isS(a)?a-32:a)
#define toC(a) a&15
 
vector <ll> sprime;
ll d[100005];
 
vector<char> prime (1000000+5, true);
void sieve(ll n)
{   
    //O(nloglogn) time
    prime[0] = prime[1] = false;
    for (ll i=2; i<=n; ++i)
    {
        if (prime[i])
        {
            sprime.pb(i);           
            for (ll j=2*i; j<=n; j+=i)
            {
                prime[j] = false;
            }
        }
    }
}
#define forn(i,e) for(int i = 0; i < e; i++)
#define rep(i,a,b) for(i=a;i<b;i++)
#define REP(i,a,b) for(i=a;i<=b;i++)
#define reprev(i,a,b) for(i=a;i>=b;i--)
 
// Convex Hull Trick
typedef ll LT;
typedef pair<LT,LT> LP;
struct CHT {
  deque<LP> d;
  LT get(LT x) {
    while (sz(d) >= 2) {
      LT a = d[0].fi*x + d[0].se;
      LT b = d[1].fi*x + d[1].se;
      if (a <= b) break; // get min
      d.pop_front();
    }
    return d[0].fi*x + d[0].se;
  }
  void add(LP x) { // decreasing order of x.fi
    while (sz(d) >= 2) {
      LP y = d[sz(d) - 1];
      LP z = d[sz(d) - 2];
      if ((x.se-y.se)*(z.fi-y.fi) >= (y.fi-x.fi)*(y.se-z.se)) break; // both
      d.pop_back();
    }
    d.pb(x);
  }
};
//
 
#define leadzero(a) __builtin_clz(a) //count leading zeros
#define trailzero(a) __builtin_ctz(a) //count trailing zeros
#define bitcount(a) __builtin_popcount(a) // count set bits (add ll)
#define ln cout<<endl
#define sp cout<<" "
#define maxe(a) *max_element(a.begin(),a.end())
#define maxi(a) max_element(a.begin(),a.end())-a.begin()
#define mine(a) *min_element(a.begin(),a.end())
#define mini(a) min_element(a.begin(),a.end())-a.begin()
ll gcd(ll a, ll b) { if(b == 0) return a; return gcd(b, a % b); }
ll fastpowMOD(ll a, ll p,ll MOD){ if(p==0) return 1; ll z = fastpowMOD(a,p/2,MOD); z = (z*z)%MOD; if(p%2) z = (z*a)%MOD; return z; }
ll fastpow(ll a, ll p){ if(p==0) return 1; ll z = fastpow(a,p/2); z = (z*z); if(p%2) z = (z*a); return z; }
//CHT stands for convex hull trick
ll search1(ll a[], ll f,ll n)
{
    ll left=1;
    ll right=n+1;
    while(right-left>1)
    {
        ll mid = left + (right-left)/2;
        if(a[mid]<f)
        {
            left=mid;
        }
        else right=mid;
    }
    return a[left];
}
 
const ll INF = 1e18;
vector<ll> lazy;
vector<ll> tree;
 
void updateRange(ll node, ll start, ll end, ll l, ll r, ll val)
{
    if(lazy[node] != 0)
    { 
        // This node needs to be updated
        tree[node] += (/*end - start*/ + 1) * lazy[node];    // Update it
        if(start != end)
        {
            lazy[node*2] += lazy[node];                  // Mark child as lazy
            lazy[node*2+1] += lazy[node];                // Mark child as lazy
        }
        lazy[node] = 0;                                  // Reset it
    }
    if(start > end or start > r or end < l)              // Current segment is not within range [l, r]
        return;
    if(start >= l and end <= r)
    {
        // Segment is fully within range
        lazy[node] += (/*end - start*/ + 1) * val;
        if(start != end)
        {
            // Not leaf node
            lazy[node*2] += lazy[node];
            lazy[node*2+1] += lazy[node];
        }
        tree[node]+=lazy[node];
        lazy[node]=0;
        return;
    }
    ll mid = (start + end) / 2;
    updateRange(node*2, start, mid, l, r, val);        // Updating left child
    updateRange(node*2 + 1, mid + 1, end, l, r, val);   // Updating right child
    tree[node] = max(tree[node*2],tree[node*2+1]);        // Updating root with max value 
}
 
ll queryRange(ll node, ll start, ll end, ll l, ll r)
{
    if(lazy[node] != 0)
    {
        // This node needs to be updated
        tree[node] += (/*end - start +*/ 1) * lazy[node];            // Update it
        if(start != end)
        {
            lazy[node*2] += lazy[node];         // Mark child as lazy
            lazy[node*2+1] += lazy[node];    // Mark child as lazy
        }
        lazy[node] = 0;                 // Reset it
    }
 
    if(start > end or start > r or end < l)
        return -INF;         // Out of range
 
    if(start >= l and end <= r)             // Current segment is totally within range [l, r]
        return tree[node];
    ll mid = (start + end) / 2;
    if(r<=mid)return queryRange(node*2, start, mid, l, r);
    else if(l>=mid+1) return queryRange(node*2 + 1, mid + 1, end, l, r);
    ll p1 = queryRange(node*2, start, mid, l, r);         // Query left child
    ll p2 = queryRange(node*2 + 1, mid + 1, end, l, r); // Query right child
    //tree[node] = max(tree[node*2],tree[node*2+1]);
    return max(p1,p2);
}
void convert_reduce(ll arr[], ll n) 
{ 
    vector <pair<ll, ll> > v; 
    for (ll i = 0; i < n; i++)v.push_back(make_pair(arr[i], i)); 
    sort(v.begin(), v.end());
    ll cnt = 0; 
    for (ll i=0; i<n; i++)
    {
        arr[v[i].second] = cnt;
        ll j = i;
        while(j<n && v[j].fi==v[i].fi)arr[v[j++].second] = cnt;
        i=j-1;
        cnt++;
    } 
} 

//for bfs 
ll dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};

bool valid(ll i, ll j,ll n,ll m) { return i >= 1 && i <= n && j >= 1 && j <= m; }

ll MOD = 998244353;

/*
D - Knapsack Queries on a tree
https://atcoder.jp/contests/tokiomarine2020/tasks/tokiomarine2020_d
https://img.atcoder.jp/tokiomarine2020/editorial.pdf

For simplicity, let N = 2^(2K) −1. Also, let M := max Li. We will 
introduce an algorithm that works in O(N +2^K(M +Q)) time. In this 
problem, K = 9 and M = 105, so 2^K(M +Q) ≈ 10^8, and thus it will 
run in time.
First, for each vertex v at depth K −1 or shallower and each 
non-negative integer k at most M, we compute Dv,k: the maximum possible 
total value of items when we choose items from v and its ancestors so 
that the total weight is at most k. By computing them in the order
from distance from the root, we can find all Dv,k in O(2KM) time.
Then, we will process the query. Let v be the specified vertex. 
Among the vertices that are v or its ancestors, at most K have 
depths of K or deeper. Thus, we can try all subsets of the
items in the vertices at depth K or deeper in O(2K) time. After 
deciding the subset to choose from the vertices at depth K or deeper, 
we can use D∗,∗ to know the optimal subset to choose
from the vertices at depth K − 1 or shallower in O(1) time. 
Thus, we can process each query in O(2^K) time.
Therefore, we can solve the problem in a total of O(N + 2^K(M + Q)) time.

*/
const ll N = (ll)3e5+5;
ll v[N],w[N];
ll dp[(1<<9)+5][(ll)1e5+4];

int main(){
    //fastio;
    ll n,i,j,k,q;
    //cin>>n;
    scanf("%lld",&n);
    REP(i,1,n)scanf("%lld %lld",&v[i],&w[i]);
    ll lim = (1<<9);
    //ll dp[lim+5][(ll)1e5];
    memset(dp,0,sizeof(dp));
    rep(i,1,min(lim,n+1))
    {
        ll par = i/2;
        rep(j,0,(ll)1e5+1)
        {
            dp[i][j]=dp[par][j];
            if(j>=w[i])
            {
                dp[i][j]=max(dp[i][j],dp[par][j-w[i]]+v[i]);
            }
            if(j>0)dp[i][j]=max(dp[i][j],dp[i][j-1]);
        }
    }
    scanf("%lld",&q);
    ll val[11];
    REP(i,0,10)val[i]=(1<<i);
    while(q--)
    {
        ll u,L;
        scanf("%lld %lld",&u,&L);
        if(u<lim)
        {
            printf("%lld\n",dp[u][L]);
            continue;
        }
        vl path;
        ll tm = u;
        while(tm>0)
        {
            path.pb(tm);
            tm/=2;
        }
        ll len = sz(path)-9;
        ll ans=0;
        rep(i,0,val[len])
        {
            ll sumv=0,sumw=0;
            rep(j,0,len)
            {
                if(i >> j & 1)
                {
                    sumv+=v[path[j]];
                    sumw+=w[path[j]];
                }
            }
            if(sumw<=L)
            {
                ans = max(ans,sumv+dp[path[len]][L-sumw]);
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}