1312_G. Autocompletion
You are given a set of strings S. Each string consists of lowercase Latin letters.

For each string in this set, you want to calculate the minimum number of seconds required to type this string. To type a string, you have to start with an empty string and transform it into the string you want to type using the following actions:

  * if the current string is t, choose some lowercase Latin letter c and append it to the back of t, so the current string becomes t + c. This action takes 1 second; 
  * use autocompletion. When you try to autocomplete the current string t, a list of all strings s ∈ S such that t is a prefix of s is shown to you. This list includes t itself, if t is a string from S, and the strings are ordered lexicographically. You can transform t into the i-th string from this list in i seconds. Note that you may choose any string from this list you want, it is not necessarily the string you are trying to type. 



What is the minimum number of seconds that you have to spend to type each string from S?

Note that the strings from S are given in an unusual way.

Input

The first line contains one integer n (1 ≤ n ≤ 10^6).

Then n lines follow, the i-th line contains one integer p_i (0 ≤ p_i < i) and one lowercase Latin character c_i. These lines form some set of strings such that S is its subset as follows: there are n + 1 strings, numbered from 0 to n; the 0-th string is an empty string, and the i-th string (i ≥ 1) is the result of appending the character c_i to the string p_i. It is guaranteed that all these strings are distinct.

The next line contains one integer k (1 ≤ k ≤ n) — the number of strings in S.

The last line contains k integers a_1, a_2, ..., a_k (1 ≤ a_i ≤ n, all a_i are pairwise distinct) denoting the indices of the strings generated by above-mentioned process that form the set S — formally, if we denote the i-th generated string as s_i, then S = {s_{a_1}, s_{a_2}, ..., s_{a_k}}.

Output

Print k integers, the i-th of them should be equal to the minimum number of seconds required to type the string s_{a_i}.

Examples

Input


10
0 i
1 q
2 g
0 k
1 e
5 r
4 m
5 h
3 p
3 e
5
8 9 1 10 6


Output


2 4 1 3 3 


Input


8
0 a
1 b
2 a
2 b
4 a
4 b
5 c
6 d
5
2 3 4 7 8


Output


1 2 2 4 4 

Note

In the first example, S consists of the following strings: ieh, iqgp, i, iqge, ier.
```python
import io
import os

DEBUG = False


def dfs(trie, root, preorder=None, postorder=None):
    stack = [root]
    seen = set()
    while stack:
        nodeId = stack.pop()
        if nodeId not in seen:
            if preorder:
                preorder(nodeId)
            stack.append(nodeId)
            seen.add(nodeId)
            for c, childId in reversed(trie[nodeId]):
                stack.append(childId)
        else:
            if postorder:
                postorder(nodeId)


def solve(N, PC, K, A):
    ROOT = 0
    trie = {ROOT: []}  # nodeId to list of (character, nodeId)
    parent = {}
    for i, (p, c) in enumerate(PC, 1):  # i starts from 1
        trie[p].append((c, i))
        assert i not in trie
        trie[i] = []
        parent[i] = p

    terminal = set(A)

    # Sort children of each node by character
    for children in trie.values():
        children.sort()

    # DFS
    offset = 0
    ancestor = []
    dist = {}

    def getDistPre(nodeId):
        nonlocal offset

        best = None
        if nodeId != 0:
            assert nodeId in parent
            best = 1 + dist[parent[nodeId]]

            if nodeId in terminal:
                best = min(best, ancestor[-1] + offset + 1)
            ancestor.append(min(ancestor[-1], best - offset))

            if nodeId in terminal:
                offset += 1
        else:
            # Is root
            best = 0
            ancestor.append(0)
        dist[nodeId] = best

    def getDistPost(nodeId):
        ancestor.pop()

    dfs(trie, ROOT, preorder=getDistPre, postorder=getDistPost)

    if DEBUG:

        def printNode(nodeId, word):
            return (
                str(nodeId)
                + "\t"
                + word
                + ("$" if nodeId in terminal else "")
                + "\t"
                + "dist: "
                + str(dist[nodeId])
            )
            return str(nodeId) + "\t" + word + ("$" if nodeId in terminal else "")

        def printGraph(nodeId, path):
            W = 8
            depth = len(path)
            for ch, childId in trie[nodeId]:
                path.append(ch)
                print(
                    (
                        " " * (W * depth)
                        + "└"
                        + ch.center(W - 1, "─")
                        + str(childId)
                        + ("$" if childId in terminal else "")
                    ).ljust(50)
                    + printNode(childId, "".join(path))
                )
                printGraph(childId, path)
                path.pop()

        printGraph(ROOT, [])

    out = []
    for a in A:
        out.append(str(dist[a]))
    return " ".join(out)


if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    N, = list(map(int, input().split()))
    PC = []
    for i in range(N):
        p, c = input().decode().split()
        PC.append((int(p), str(c)))
    K, = list(map(int, input().split()))
    A = list(map(int, input().split()))
    ans = solve(N, PC, K, A)
    print(ans)

```