#include <bits/stdc++.h>
using namespace std;
char *fs, *ft, buf[1 << 15];
inline char getc() {
  return (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 15, stdin), fs == ft))
             ? 0
             : *fs++;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getc();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getc();
  }
  return x * f;
}
const int MAXN = 100010;
int n, len, mx, tot, top, m;
int a[MAXN], s[MAXN], cnt[MAXN], root[MAXN], sz[MAXN], son[MAXN];
int lin[MAXN], nex[MAXN], ver[MAXN], d[MAXN], ans[MAXN];
vector<pair<long long, long long> > g[MAXN], w[MAXN];
inline void add(int x, int y) {
  ver[++len] = y;
  nex[len] = lin[x];
  lin[x] = len;
}
inline void dfs(int x, int father) {
  d[x] = d[father] + 1;
  sz[x] = 1;
  s[++top] = x;
  for (int i = 0; i < g[x].size(); ++i) {
    int tn = g[x][i].first;
    if (tn > d[father]) continue;
    int ss = s[top - tn];
    w[ss].push_back(make_pair(d[x], g[x][i].second));
  }
  for (int i = lin[x]; i; i = nex[i]) {
    int tn = ver[i];
    if (tn == father) continue;
    dfs(tn, x);
    sz[x] += sz[tn];
    if (sz[son[x]] < sz[tn]) son[x] = tn;
  }
  --top;
}
inline void update(int x, int father, int op, int target) {
  cnt[d[x]] += op;
  for (int i = lin[x]; i; i = nex[i]) {
    int tn = ver[i];
    if (tn == father || tn == target) continue;
    update(tn, x, op, target);
  }
}
inline void dfs(int x, int father, int op) {
  for (int i = lin[x]; i; i = nex[i]) {
    int tn = ver[i];
    if (tn == father || tn == son[x]) continue;
    dfs(tn, x, 0);
  }
  if (son[x]) dfs(son[x], x, 1);
  update(x, father, 1, son[x]);
  for (int i = 0; i < w[x].size(); ++i)
    ans[w[x][i].second] += cnt[w[x][i].first];
  if (!op) update(x, father, -1, 0);
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    int x = read();
    if (!x)
      root[++tot] = i;
    else
      add(x, i);
  }
  m = read();
  for (int i = 1; i <= m; ++i) {
    int x, y;
    x = read();
    y = read();
    g[x].push_back(make_pair(y, i));
  }
  for (int i = 1; i <= tot; ++i) dfs(root[i], 0);
  for (int i = 1; i <= tot; ++i) dfs(root[i], 0, 0);
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i] ? ans[i] - 1 : 0);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

Polycarpus got hold of a family relationship tree. The tree describes family relationships of n people, numbered 1 through n. Each person in the tree has no more than one parent.

Let's call person a a 1-ancestor of person b, if a is the parent of b.

Let's call person a a k-ancestor (k > 1) of person b, if person b has a 1-ancestor, and a is a (k - 1)-ancestor of b's 1-ancestor. 

Family relationships don't form cycles in the found tree. In other words, there is no person who is his own ancestor, directly or indirectly (that is, who is an x-ancestor for himself, for some x, x > 0).

Let's call two people x and y (x ≠ y) p-th cousins (p > 0), if there is person z, who is a p-ancestor of x and a p-ancestor of y.

Polycarpus wonders how many counsins and what kinds of them everybody has. He took a piece of paper and wrote m pairs of integers vi, pi. Help him to calculate the number of pi-th cousins that person vi has, for each pair vi, pi.

Input

The first input line contains a single integer n (1 ≤ n ≤ 105) — the number of people in the tree. The next line contains n space-separated integers r1, r2, ..., rn, where ri (1 ≤ ri ≤ n) is the number of person i's parent or 0, if person i has no parent. It is guaranteed that family relationships don't form cycles.

The third line contains a single number m (1 ≤ m ≤ 105) — the number of family relationship queries Polycarus has. Next m lines contain pairs of space-separated integers. The i-th line contains numbers vi, pi (1 ≤ vi, pi ≤ n).

Output

Print m space-separated integers — the answers to Polycarpus' queries. Print the answers to the queries in the order, in which the queries occur in the input.

Examples

Input

6
0 1 1 0 4 4
7
1 1
1 2
2 1
2 2
4 1
5 1
6 1


Output

0 0 1 0 0 1 1 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
char *fs, *ft, buf[1 << 15];
inline char getc() {
  return (fs == ft &&
          (ft = (fs = buf) + fread(buf, 1, 1 << 15, stdin), fs == ft))
             ? 0
             : *fs++;
}
inline int read() {
  int x = 0, f = 1;
  char ch = getc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getc();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getc();
  }
  return x * f;
}
const int MAXN = 100010;
int n, len, mx, tot, top, m;
int a[MAXN], s[MAXN], cnt[MAXN], root[MAXN], sz[MAXN], son[MAXN];
int lin[MAXN], nex[MAXN], ver[MAXN], d[MAXN], ans[MAXN];
vector<pair<long long, long long> > g[MAXN], w[MAXN];
inline void add(int x, int y) {
  ver[++len] = y;
  nex[len] = lin[x];
  lin[x] = len;
}
inline void dfs(int x, int father) {
  d[x] = d[father] + 1;
  sz[x] = 1;
  s[++top] = x;
  for (int i = 0; i < g[x].size(); ++i) {
    int tn = g[x][i].first;
    if (tn > d[father]) continue;
    int ss = s[top - tn];
    w[ss].push_back(make_pair(d[x], g[x][i].second));
  }
  for (int i = lin[x]; i; i = nex[i]) {
    int tn = ver[i];
    if (tn == father) continue;
    dfs(tn, x);
    sz[x] += sz[tn];
    if (sz[son[x]] < sz[tn]) son[x] = tn;
  }
  --top;
}
inline void update(int x, int father, int op, int target) {
  cnt[d[x]] += op;
  for (int i = lin[x]; i; i = nex[i]) {
    int tn = ver[i];
    if (tn == father || tn == target) continue;
    update(tn, x, op, target);
  }
}
inline void dfs(int x, int father, int op) {
  for (int i = lin[x]; i; i = nex[i]) {
    int tn = ver[i];
    if (tn == father || tn == son[x]) continue;
    dfs(tn, x, 0);
  }
  if (son[x]) dfs(son[x], x, 1);
  update(x, father, 1, son[x]);
  for (int i = 0; i < w[x].size(); ++i)
    ans[w[x][i].second] += cnt[w[x][i].first];
  if (!op) update(x, father, -1, 0);
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) {
    int x = read();
    if (!x)
      root[++tot] = i;
    else
      add(x, i);
  }
  m = read();
  for (int i = 1; i <= m; ++i) {
    int x, y;
    x = read();
    y = read();
    g[x].push_back(make_pair(y, i));
  }
  for (int i = 1; i <= tot; ++i) dfs(root[i], 0);
  for (int i = 1; i <= tot; ++i) dfs(root[i], 0, 0);
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i] ? ans[i] - 1 : 0);
  return 0;
}

```