/*
    问题代码
    思路：将问题转换为求链表的第一个公共子节点
*/
#include<iostream>
#include<cstring>
#include<unordered_map>
using namespace std;
int n, m;

const int N = 10010;
int in[N], pre[N];
int l[N], r[N];

unordered_map<int, int> pos, father;
unordered_map<int, bool> st, mark;
//bool mark[N];  //问题代码：无法标记负数，可以用unordered_map代替

int build(int pl, int pr, int il, int ir){
    if(pl > pr) return 0;
    int u = pre[pl];
    int k = pos[u];

    int num = k - il;
    l[u] = build(pl + 1, pl + num, il, k - 1);
    r[u] = build(pl + num + 1, pr, k + 1, ir);
    return u;
}

void dfs(int u){
    if(!u) return;
    father[l[u]] = u, dfs(l[u]);
    father[r[u]] = u, dfs(r[u]);
}
int find(int &a, int &b, int root){
    int t1 = a, t2 = b;
    if(t1 == t2) return t1;

    while(t1){
        mark[t1] = true;
        if(t1 == root) break;
        t1 = father[t1];
    }
    while(t2){
        if(mark[t2]) return t2;
        t2 = father[t2];
    }
}

int main(){
    cin >> m >> n;
    for(int i = 0; i < n; i++ ) {cin >> in[i]; pos[in[i]] = i;}
    for(int i = 0; i < n; i++) {cin >> pre[i]; st[pre[i]] = true;}

    int root = build(0, n - 1, 0, n - 1);
    dfs(root);

    for(int j = 0; j < m ; j ++){
        int a , b;
        cin >> a >> b;
        //memset(mark, false, sizeof mark);
        mark.clear();
        if(!st[a] && !st[b])  printf("ERROR: %d and %d are not found.\n", a, b);
        else if(!st[a])  printf("ERROR: %d is not found.\n", a);
        else if(!st[b])  printf("ERROR: %d is not found.\n", b);
        else{

            int t = find(a, b, root);
            if(t != b && t != a) printf("LCA of %d and %d is %d.\n", a, b, t);
            else if(t == b && t != a) printf("%d is an ancestor of %d.\n", b, a);
            else if(t != b && t == a) printf("%d is an ancestor of %d.\n", a, b);
            else  printf("%d is an ancestor of %d.\n", a, b);
        }
    }
    return 0;
}