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 (Clang 10.0.0) (judge verdict: Accepted)

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

#define MAX_N  ((1 << 18) - 1)
#define MAX_L  100000

#define min(a, b)  ((a) < (b) ? (a) : (b))
#define max(a, b)  ((a) > (b) ? (a) : (b))

int n;
int val[MAX_N], wgt[MAX_N];

int dpn;

typedef struct wgtval {
  int wgt, val;
} wgtval;

struct wgtval_list {
  int size;
  wgtval *list;
} dp[MAX_N];

#define set_wgtval(wv, w, v)  ((wv).wgt = (w), (wv).val = (v))

int compare_wgt(const void *a, const void *b)
{
  const wgtval *p = a;
  const wgtval *q = b;
  if (p->wgt == q->wgt) {
    return p->val < q->val ? -1 :
           p->val > q->val ?  1 : 0;
  }
  return p->wgt < q->wgt ? -1 : 1;
}

void compute_dp()
{
  int u;

  /* initialize */
  dpn = min(n, 1 << 12);
  dp[0].size = 2;
  dp[0].list = malloc(2 * sizeof(wgtval));
  set_wgtval(dp[0].list[0], 0, 0);
  set_wgtval(dp[0].list[1], wgt[0], val[0]);

  /* compute the dp table */
  for (u = 1; u < dpn; u++) {
    int parent = (u - 1) / 2;
    int tail, i;

    dp[u].size = dp[parent].size * 2;
    dp[u].list = malloc(dp[u].size * sizeof(wgtval));
    for (i = 0; i < dp[parent].size; i++) {
      dp[u].list[i * 2] = dp[parent].list[i];
      set_wgtval(dp[u].list[i * 2 + 1],
                 dp[parent].list[i].wgt + wgt[u],
                 dp[parent].list[i].val + val[u]);
    }
    qsort(dp[u].list, dp[u].size, sizeof(wgtval), compare_wgt);

    /* remove unnecessary entries */
    tail = 0;
    for (i = 1; i < dp[u].size; i++) {
      if (dp[u].list[i].wgt > MAX_L) {
        break;
      }
      if (dp[u].list[tail].wgt == dp[u].list[i].wgt) {
        dp[u].list[tail] = dp[u].list[i];
      } else if (dp[u].list[tail].val < dp[u].list[i].val) {
        tail++;
        dp[u].list[tail] = dp[u].list[i];
      }
    }
    dp[u].size = tail + 1;
  }
}

int bsearch_right(wgtval *list, int size, int target)
{
  int lo = 0;
  int hi = size;
  while (lo < hi) {
    int mid = (lo + hi) / 2;
    if (target < list[mid].wgt) {
      hi = mid;
    } else {
      lo = mid + 1;
    }
  }
  return lo;
}

int solve(int u, int l)
{
  int parent;
  int s;

  if (u < dpn) {
    int x = bsearch_right(dp[u].list, dp[u].size, l) - 1;
    return dp[u].list[x].val;
  }

  parent = (u - 1) / 2;
  s = solve(parent, l);
  if (l >= wgt[u]) {
    int t = val[u] + solve(parent, l - wgt[u]);
    s = max(s, t);
  }
  return s;
}

int main()
{
  int i, q;

  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d%d", &val[i], &wgt[i]);
  }

  compute_dp();

  scanf("%d", &q);
  for (i = 0; i < q; i++) {
    int u, l;
    scanf("%d%d", &u, &l);
    printf("%d\n", solve(u - 1, l));
  }

  return 0;
}