#include <bits/stdc++.h>
using namespace std;
const int MAXN = 40000;
const int MAXM = 40000;
int n, q, root;          // 点数、询问个数、根节点
vector<int> e[MAXN + 5]; // 存树
int fa[MAXN + 5];
int maxDep, dep[MAXN + 5];
void dfs(int u, int father)
{
    fa[u] = father;
    for (int i = 0; i < e[u].size(); i++)
    {
        int v = e[u][i];
        if (v == father)
            continue;
        dep[v] = dep[u] + 1;
        maxDep = max(maxDep, dep[v]);
        dfs(v, u);
    }
}
int lg2[MAXN + 5]; // 替代 log2(i)
int dp[MAXN][35];  // i号点的 2^j 层祖先
int lca(int u, int v)
{
    if (dep[u] > dep[v])
        swap(u, v); // 让u的高度不低于v
    if (u == v)
        return u;
    // 把 v 提到与 u 同样高度
    for (int j = lg2[n]; j >= 0; j--)
        if ((1 << j) <= dep[v] - dep[u])
        {
            v = dp[v][j];
        }
    if (u == v)
        return u;
    // 找lca
    for (int j = lg2[n]; j >= 0; j--)
        if (dp[v][j] != dp[u][j])
        {
            v = dp[v][j],
            u = dp[u][j];
        }
    return dp[u][0];
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //  输入
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int u, v;
        cin >> u >> v;
        fa[u] = fa[v] = -1;
        if (v == -1)
            root = u;
        else
        {
            e[u].push_back(v);
            e[v].push_back(u);
        }
    }
    // 预处理：最大层数 maxDep、每个点的深度、记录父节点、log2[]、dp数组
    dep[root] = 1;
    maxDep = 1;
    dfs(root, 0);
    lg2[1] = 0;
    for (int i = 2; i <= n; i++)
        // i == 2^(log2(i-1)+1)
        if (i == (1 << (lg2[i - 1] + 1)))
            lg2[i] = lg2[i - 1] + 1;
        else
            lg2[i] = lg2[i - 1];
    for (int j = 0; j <= lg2[n]; j++)
    {
        for (int i = 1; i <= MAXN; i++)
        {
            if (fa[i] == -1)
                continue;
            // 求解 dp[i][j]
            if (j == 0)
                dp[i][j] = fa[i];
            else
                dp[i][j] = dp[dp[i][j - 1]][j - 1];
        }
    }
    // q 次询问
    cin >> q;
    while (q--)
    {
        int u, v;
        cin >> u >> v; // 求 lca(u,v)
        int x = lca(u, v);
        if (x == u)
            cout << 1 << "\n";
        else if (x == v)
            cout << 2 << "\n";
        else
            cout << 0 << "\n";
    }
    return 0;
}