Polycarpus got hold of a family tree. The found tree describes the family relations of n people, numbered from 1 to n. Every person in this tree has at most one direct ancestor. Also, each person in the tree has a name, the names are not necessarily unique.

We call the man with a number a a 1-ancestor of the man with a number b, if the man with a number a is a direct ancestor of the man with a number b.

We call the man with a number a a k-ancestor (k > 1) of the man with a number b, if the man with a number b has a 1-ancestor, and the man with a number a is a (k - 1)-ancestor of the 1-ancestor of the man with a number b.

In the tree the family ties do not form cycles. In other words there isn't a person who is his own direct or indirect ancestor (that is, who is an x-ancestor of himself, for some x, x > 0).

We call a man with a number a the k-son of the man with a number b, if the man with a number b is a k-ancestor of the man with a number a.

Polycarpus is very much interested in how many sons and which sons each person has. He took a piece of paper and wrote m pairs of numbers vi, ki. Help him to learn for each pair vi, ki the number of distinct names among all names of the ki-sons of the man with number vi.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. Next n lines contain the description of people in the tree. The i-th line contains space-separated string si and integer ri (0 ≤ ri ≤ n), where si is the name of the man with a number i, and ri is either the number of the direct ancestor of the man with a number i or 0, if the man with a number i has no direct ancestor. 

The next line contains a single integer m (1 ≤ m ≤ 105) — the number of Polycarpus's records. Next m lines contain space-separated pairs of integers. The i-th line contains integers vi, ki (1 ≤ vi, ki ≤ n).

It is guaranteed that the family relationships do not form cycles. The names of all people are non-empty strings, consisting of no more than 20 lowercase English letters.

Output

Print m whitespace-separated integers — the answers to Polycarpus's records. Print the answers to the records in the order, in which the records occur in the input.

Examples

Input

6
pasha 0
gerald 1
gerald 1
valera 2
igor 3
olesya 1
5
1 1
1 2
1 3
3 1
6 1


Output

2
2
0
1
0


Input

6
valera 0
valera 1
valera 1
gerald 0
valera 4
kolya 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1


Output

1
0
0
0
2
0
0
n = int(raw_input())
child = [set({}) for i in range(0, n + 1)]
query = [list({}) for i in range(0, n + 1)]
ans = [dict(set({})) for i in range(0, n + 1)]
ss = [0 for i in range(0, n + 1)]
fa = [0 for i in range(0, n + 1)]
hash_string = dict()
nowstrings = 0
for i in range(1, n + 1):
    s = raw_input().split()
    child[int(s[1])] |= {i}
    fa[i] = int(s[1])
    if s[0] not in hash_string:
        nowstrings += 1
        hash_string[s[0]] = nowstrings
    ss[i] = hash_string[s[0]]
fa[0] = -1
Child = [list(child[i]) for i in range(0, n+1)]
lenChild = [len(Child[i]) for i in range(0, n+1)]
nowChild = [0 for i in range(0, n+1)]
m = int(raw_input())
answer = [0 for i in range(0, m)]
for i in range(0, m):
    s = raw_input().split()
    query[int(s[0])] += {int(s[1])}
    query[int(s[0])] += {i}
def dfs(x, dep):
    while x != -1:
        if nowChild[x] == lenChild[x]:
            if x != 0:
                ans[x][dep] = {ss[x]}
                for nowlistindex in range(0, len(query[x]), 2):
                    i = query[x][nowlistindex]
                    j = query[x][nowlistindex+1]
                    if dep + i not in ans[x]:
                        answer[j] = 0
                    else:
                        answer[j] = len(ans[x][dep + i])
                """count ans"""
                if len(ans[x]) < len(ans[fa[x]]):
                    for j in ans[x]:
                        if j not in ans[fa[x]]:
                            ans[fa[x]][j] = ans[x][j]
                        elif len(ans[x][j]) < len(ans[fa[x]][j]):
                            ans[fa[x]][j] |= ans[x][j]
                            ans[x][j].clear()
                        else:
                            ans[x][j] |= ans[fa[x]][j]
                            ans[fa[x]][j].clear()
                            ans[fa[x]][j] = ans[x][j]
                    ans[x].clear()
                else:
                    for j in ans[fa[x]]:
                        if j not in ans[x]:
                            ans[x][j] = ans[fa[x]][j]
                        elif len(ans[x][j]) > len(ans[fa[x]][j]):
                            ans[x][j] |= ans[fa[x]][j]
                            ans[fa[x]][j].clear()
                        else:
                            ans[fa[x]][j] |= ans[x][j]
                            ans[x][j].clear()
                            ans[x][j] = ans[fa[x]][j]
                    ans[fa[x]].clear()
                    ans[fa[x]] = ans[x]
            x = fa[x]
            nowChild[x] += 1
            dep -= 1
        else:
            x = Child[x][nowChild[x]]
            dep += 1
dfs(0, 0)
for i in range(0, m):
    print(answer[i])
