The lowest common ancestor (LCA) of two nodes U and V in a tree is the deepest node that has both U and V as descendants.

A binary search tree (BST) is recursively defined as a binary tree which has the following properties:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
• Both the left and right subtrees must also be binary search trees.

Given any two nodes in a BST, you are supposed to find their LCA.

Input Specification:

Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 1,000), the number of pairs of nodes to be tested; and N (≤ 10,000), the number of keys in the BST, respectively. In the second line, N distinct integers are given as the preorder traversal sequence of the BST. Then M lines follow, each contains a pair of integer keys U and V. All the keys are in the range of int.

Output Specification:

For each given pair of U and V, print in a line LCA of U and V is A. if the LCA is found and A is the key. But if A is one of U and V, print X is an ancestor of Y. where Xis A and Y is the other node. If U or V is not found in the BST, print in a line ERROR: U is not found. or ERROR: V is not found. or ERROR: U and V are not found..

Sample Input:

6 8

6 3 1 2 5 4 8 7

2 5

8 7

1 9

12 -3

0 8

99 99

Sample Output:

LCA of 2 and 5 is 3.

8 is an ancestor of 7.

ERROR: 9 is not found.

ERROR: 12 and -3 are not found.

ERROR: 0 is not found.

ERROR: 99 and 99 are not found.

#include <stdio.h>

#include <algorithm>

#include <iostream>

#include <map>

#include <vector>

#include <set>

using namespace std;

const int maxn=;

int n,m,k;

int inorder[maxn],preorder[maxn];

struct node{

    int data;

    node* left;

    node* right;

};

node* create(int prel,int prer,int inl,int inr){

    if(prel>prer){

        return NULL;

    }

    node* root=new node;

    root->data=preorder[prel];

    int k;

    for(k=inl;k<=inr;k++){

        if(inorder[k]==preorder[prel]){

            break;

        }

    }

    int numleft=k-inl;

    root->left = create(prel+,prel+numleft,inl,k-);

    root->right = create(prel+numleft+,prer,k+,inr);

    return root;

}

bool findnode(int x){

    for(int i=;i<m;i++){

        if(preorder[i]==x) return true;

    }

    return false;

}

node* lca(node* root,int x1,int x2){

    if(root==NULL)return NULL;

    if(root->data==x1 || root->data==x2) return root;

    node* left = lca(root->left,x1,x2);

    node* right = lca(root->right,x1,x2);

    if(left && right) return root;

    else if(left==NULL) return right;

    else return left;

}

int main(){

    scanf("%d %d",&n,&m);

    for(int i=;i<m;i++){

        int c1;

        scanf("%d",&c1);

        preorder[i]=c1;

        inorder[i]=c1;

    }

    sort(inorder,inorder+m);

    node *root=create(,m-,,m-);

    for(int i=;i<n;i++){

        int x1,x2;

        scanf("%d %d",&x1,&x2);

        if(!findnode(x1) && !findnode(x2)){

            printf("ERROR: %d and %d are not found.\n",x1,x2);

        }

        else if(!findnode(x1)) printf("ERROR: %d is not found.\n",x1);

        else if(!findnode(x2)) printf("ERROR: %d is not found.\n",x2);

        else{

            node* res = lca(root,x1,x2);

            if(res->data == x1) printf("%d is an ancestor of %d.\n",x1,x2);

            else if(res->data == x2) printf("%d is an ancestor of %d.\n",x2,x1);

            else printf("LCA of %d and %d is %d.\n",x1,x2,res->data);

        }

    }

}