Here

• 利用回文串，从左往右与从右往左的hash值相同来判断
• 从左往右，例：
• 从右往左，例：
• 由于在树上，考虑建两颗树，一颗根为最高位（up），一棵根为最低位（down)
• 
• 考虑bccb(1,2,3,4)这个
• 
• ，即bcc
• ，即b
• 
• 所以设在树上两点
• 
• (乘inv使x为最低位)
• 
• 
• 
• ，swap(u,v)即可
• 为保证hash的正确性，采用双哈希，

import java.io.*;
import java.math.BigInteger;
import java.util.*;//implements Runnable
public class Main {static long md1=(long)1e9+7;static long md2=(long)1e9+9;static long Linf=Long.MAX_VALUE/2;static int inf=Integer.MAX_VALUE/2;static int N=200010;static int n=0;static int m=0;staticclass M{long x,y;public M(){};public M(long u,long v){x=u;y=v;}}static M mul(M a,M b){return new M(a.x*b.x%md1,a.y*b.y%md2);}static M add(M a,M b){return new M((a.x+b.x)%md1,(a.y+b.y)%md2);}static M sub(M a,M b){return new M((a.x-b.x+md1)%md1,(a.y-b.y+md2)%md2);}static M[] inv;static M[] fac;static M base=new M(29,131);static long qm2(long a,long b,long md){long res=1;while(b>0){if((b&1)==1)res=res*a%md;a=a*a%md;b>>=1;}return res;}static void getFac(){fac[0]=new M(1,1);up[0]=new M(0,0);down[0]=new M(0,0);for(int i=1;i<=n;++i){fac[i]=mul(fac[i-1],base);}inv[n]=new M();inv[n].x=qm2(fac[n].x,md1-2,md1);inv[n].y=qm2(fac[n].y,md2-2,md2);for(int i=n-1;i>=0;i--){inv[i]=mul(inv[i+1],base);}}staticclass Edge{int fr,to,nxt;public Edge(int u,int v){fr=u;to=v;}}static Edge[] e;static int[] head;static int cnt=0;static void addEdge(int fr,int to){cnt++;e[cnt]=new Edge(fr,to);e[cnt].nxt=head[fr];head[fr]=cnt;}static int[] fa;static int[] son;static int[] siz;static int[] dep;static int[] top;static M[] down;static M[] up;static int[] a;static void dfs1(int x,int f){siz[x]=1;dep[x]=dep[f]+1;fa[x]=f;up[x]=add(mul(up[f],base),new M(a[x],a[x]));down[x]=add(down[f],mul(new M(a[x],a[x]),fac[dep[x]]));for(int i=head[x];i>0;i=e[i].nxt){int v=e[i].to;if(v==f)continue;dfs1(v,x);siz[x]+=siz[v];if(siz[son[x]]<siz[v])son[x]=v;}}static void dfs2(int x,int tp){top[x]=tp;if(son[x]!=0)dfs2(son[x],tp);for(int i=head[x];i>0;i=e[i].nxt){int v=e[i].to;if(v==fa[x]||v==son[x])continue;dfs2(v,v);}}static int LCA(int x,int y){while(top[x]!=top[y]){if(dep[top[x]]>dep[top[y]]){int t=x;x=y;y=t;}y=fa[top[y]];}if(dep[x]>dep[y]){int t=x;x=y;y=t;}return x;}static M getHash(int u,int v,int x){M ux=mul(sub(down[u],down[fa[x]]),inv[dep[x]]);M xv=sub(up[v],mul(up[x],fac[dep[v]-dep[x]]));return add(mul(ux,fac[dep[v]-dep[x]]),xv);//从右上左下}static void getPre(){fac=new M[n+1];inv=new M[n+1];fa=new int[n+1];son=new int[n+1];siz=new int[n+1];dep=new int[n+1];top=new int[n+1];head=new int[n+1];e=new Edge[2*n+10];cnt=0;down=new M[n+1];//rhaup=new M[n+1];//haa=new int[n+1];}static void solve() throws Exception{AReader input=new AReader();
//        String fileName="";
//		Scanner input=new Scanner(new FileReader(fileName));//        BufferedReader input = new BufferedReader(new FileReader(fileName));PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));String al="abcdefghijklmnopqrstuvwxyz";char[] ac=al.toCharArray();n=input.nextInt();getPre();getFac();String string=" "+input.next();char[] s=string.toCharArray();for(int i=1;i<=n;++i){a[i]=s[i]-'a'+1;}for(int i=1;i<=n;++i){int x=input.nextInt();addEdge(x,i);addEdge(i,x);}dep[0]=-1;dfs1(1,0);dfs2(1,1);m=input.nextInt();while(m>0){m--;int x=input.nextInt();int y=input.nextInt();int p=LCA(x,y);M rtol=getHash(x,y,p);M ltor=getHash(y,x,p);if(rtol.x==ltor.x&&rtol.y==ltor.y)out.println("YES");else out.println("NO");}out.flush();out.close();}public static void main(String[] args) throws Exception{solve();}//	public static final void main(String[] args) throws Exception {
//		  new Thread(null, new Tx2(), "线程名字", 1 << 27).start();
//	}
//		@Override
//		public void run() {
//			try {
//				//原本main函数的内容
//				solve();
//
//			} catch (Exception e) {
//			}
//		}staticclass AReader{BufferedReader bf;StringTokenizer st;public AReader(){bf=new BufferedReader(new InputStreamReader(System.in));st=new StringTokenizer("");}public String nextLine() throws IOException{return bf.readLine();}public String next() throws IOException{while(!st.hasMoreTokens()){st=new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException{//确定下一个token只有一个字符的时候再用return next().charAt(0);}public int nextInt() throws IOException{return Integer.parseInt(next());}public long nextLong() throws IOException{return Long.parseLong(next());}public double nextDouble() throws IOException{return Double.parseDouble(next());}public float nextFloat() throws IOException{return Float.parseFloat(next());}public byte nextByte() throws IOException{return Byte.parseByte(next());}public short nextShort() throws IOException{return Short.parseShort(next());}public BigInteger nextBigInteger() throws IOException{return new BigInteger(next());}}
}