#include <stdio.h>
#include <ctype.h>
#include <vector>
#include <algorithm>
#include <fstream>
using namespace std;

class InParser {
private:
    FILE *fin;
    char *buff;
    int sp;

    char read_ch() {
        ++sp;
        if (sp == 4096) {
            sp = 0;
            fread(buff, 1, 4096, fin);
        }
        return buff[sp];
    }

public:
    InParser(const char* nume) {
        fin = fopen(nume, "r");
        buff = new char[4096]();
        sp = 4095;
    }

    InParser& operator >> (int &n) {
        char c;
        while (!isdigit(c = read_ch()) && c != '-');
        int sgn = 1;
        if (c == '-') {
            n = 0;
            sgn = -1;
        } else {
            n = c - '0';
        }
        while (isdigit(c = read_ch())) {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }

    InParser& operator >> (long long &n) {
        char c;
        n = 0;
        while (!isdigit(c = read_ch()) && c != '-');
        long long sgn = 1;
        if (c == '-') {
            n = 0;
            sgn = -1;
        } else {
            n = c - '0';
        }
        while (isdigit(c = read_ch())) {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }
};


int N,Q;
ofstream fo("lca.out");
vector <int> D[100001];
int E[200001],k;
int AD[100001];
int PAP[100001];
int T[200001][18],nrc,p;

void euler(int v,int ad)
{
	vector <int> :: iterator it;
	E[++k]=v;
	AD[v]=ad;
	PAP[v]=k;
	for (it=D[v].begin();it!=D[v].end();it++)
	{
		euler(*it,ad+1);
		E[++k]=v;
	}
}

int query(int st, int dr)
/// returneaza varful cu adancime minima din portiunea de indici st->dr
{
	int vfrez;
	vfrez=E[st];
	for (int p=nrc;p>=0;p--)
		if (st+(1<<p)-1<=dr)
		{
			if (AD[T[st][p]]<AD[vfrez])
				vfrez=T[st][p];
			st=st+(1<<p);
		}
	return vfrez;
}

int main()
{
	InParser fi("lca.in");
	fi>>N>>Q;
	for (int i=2;i<=N;i++)
	{
		int v;
		fi>>v;
		// v este parintele lui i
		D[v].push_back(i);
	}
	/// se obtine traversarea Euler a arborelui
	k=0;
	euler(1,0);
	/// se construieste T
	nrc=0;
	p=1;
	while ((p<<1)<=k)
	{
		p=(p<<1);
		nrc++;
	}
	/// coloana 0
	for (int i=1;i<=k;i++)
		T[i][0]=E[i];
	/// alte coloane
	for (int j=1;j<=nrc;j++)
		for (int i=1;i<=k-(1<<j)+1;i++)
			if (AD[T[i][j-1]]<=AD[T[i+(1<<(j-1))][j-1]])
				T[i][j]=T[i][j-1];
			else
				T[i][j]=T[i+(1<<(j-1))][j-1];
	/// se citesc interogarile si se raspunde la ele
	for (int i=1;i<=Q;i++)
	{
		int a,b;
		fi>>a>>b;
		int st,dr;
		st=PAP[a];
		dr=PAP[b];
		if (st>dr)
			swap(st,dr);
		fo<<query(st,dr)<<"\n";
	}
	fo.close();
	return 0;
}