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 Kotlin (1.3.71) (judge verdict: Time Limit Exceeded)


import java.lang.Integer.max
import kotlin.math.log2

fun main() {
    val n = nextInt()
    val tree = Array(n) {
        nextInt() to nextInt()
    }
    val q = nextInt()
    val res = StringBuilder()
    repeat(q) {
        val (u, l) = next2Int()
        val n = log2(u.toDouble()).toInt() + 1
        val w = IntArray(n)
        val v = IntArray(n)
        var p = u
        for (i in 0 until n) {
            v[i] = tree[p - 1].first
            w[i] = tree[p - 1].second
            p /= 2
        }
        val dp = Array(n + 1) { IntArray(l + 1) }
        for (i in n - 1 downTo 0) {
            for (j in 0..l) {
                if (j < w[i]) {
                    dp[i][j] = dp[i + 1][j]
                } else {
                    dp[i][j] = max(dp[i + 1][j], dp[i + 1][j - w[i]] + v[i])
                }
            }
        }
        res.appendln(dp[0][l])
    }
    print(res.toString())
}

val sc = java.util.Scanner(System.`in`)
fun next(): String = sc.next()
fun nextInt() = sc.nextInt()
fun next2Int() = Pair(nextInt(), nextInt())
fun next3Int() = Triple(nextInt(), nextInt(), nextInt())
fun nextIntArray(size: Int) = IntArray(size) { nextInt() }
fun nextLong() = sc.nextLong()
fun nextLine(): String = sc.nextLine()
fun nextList() = nextLine().split(" ")
fun printYesNo(isYes: Boolean) = println(if (isYes) "Yes" else "No")
fun printString(action: StringBuilder.() -> Unit) = println(buildString(action))