Recently Lynyrd and Skynyrd went to a shop where Lynyrd bought a permutation p of length n, and Skynyrd bought an array a of length m, consisting of integers from 1 to n. 

Lynyrd and Skynyrd became bored, so they asked you q queries, each of which has the following form: "does the subsegment of a from the l-th to the r-th positions, inclusive, have a subsequence that is a cyclic shift of p?" Please answer the queries.

A permutation of length n is a sequence of n integers such that each integer from 1 to n appears exactly once in it.

A cyclic shift of a permutation (p_1, p_2, …, p_n) is a permutation (p_i, p_{i + 1}, …, p_{n}, p_1, p_2, …, p_{i - 1}) for some i from 1 to n. For example, a permutation (2, 1, 3) has three distinct cyclic shifts: (2, 1, 3), (1, 3, 2), (3, 2, 1).

A subsequence of a subsegment of array a from the l-th to the r-th positions, inclusive, is a sequence a_{i_1}, a_{i_2}, …, a_{i_k} for some i_1, i_2, …, i_k such that l ≤ i_1 < i_2 < … < i_k ≤ r.

Input

The first line contains three integers n, m, q (1 ≤ n, m, q ≤ 2 ⋅ 10^5) — the length of the permutation p, the length of the array a and the number of queries.

The next line contains n integers from 1 to n, where the i-th of them is the i-th element of the permutation. Each integer from 1 to n appears exactly once.

The next line contains m integers from 1 to n, the i-th of them is the i-th element of the array a.

The next q lines describe queries. The i-th of these lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ m), meaning that the i-th query is about the subsegment of the array from the l_i-th to the r_i-th positions, inclusive.

Output

Print a single string of length q, consisting of 0 and 1, the digit on the i-th positions should be 1, if the subsegment of array a from the l_i-th to the r_i-th positions, inclusive, contains a subsequence that is a cyclic shift of p, and 0 otherwise.

Examples

Input

3 6 3
2 1 3
1 2 3 1 2 3
1 5
2 6
3 5


Output

110


Input

2 4 3
2 1
1 1 2 2
1 2
2 3
3 4


Output

010

Note

In the first example the segment from the 1-st to the 5-th positions is 1, 2, 3, 1, 2. There is a subsequence 1, 3, 2 that is a cyclic shift of the permutation. The subsegment from the 2-nd to the 6-th positions also contains a subsequence 2, 1, 3 that is equal to the permutation. The subsegment from the 3-rd to the 5-th positions is 3, 1, 2, there is only one subsequence of length 3 (3, 1, 2), but it is not a cyclic shift of the permutation.

In the second example the possible cyclic shifts are 1, 2 and 2, 1. The subsegment from the 1-st to the 2-nd positions is 1, 1, its subsequences are not cyclic shifts of the permutation. The subsegment from the 2-nd to the 3-rd positions is 1, 2, it coincides with the permutation. The subsegment from the 3 to the 4 positions is 2, 2, its subsequences are not cyclic shifts of the permutation.
import java.util.*;
import java.io.*;

public class Main {
    public static void main(String args[]) {new Main().run();}

    FastReader in = new FastReader();
    PrintWriter out = new PrintWriter(System.out);
    void run() {
        work();
        out.flush();
    }
    long mod=1000000007;
    long gcd(long a,long b) {
        return a==0?b:gcd(b%a,a);
    }
    void work() {
        int n=ni(),m=ni(),q=ni();
        int[] A=nia(n);
        int[][] dp=new int[m][20];
        int[] rec=new int[n+1];
        int[] pre=new int[n+1];
        for(int i=0;i<n;i++){
            pre[A[i]]=A[(i-1+n)%n];
        }
        Arrays.fill(rec,-1);
        int[] B=nia(m);
        int[] rec2=new int[m];
        for(int i=0,d=-1;i<m;i++){
            rec[B[i]]=i;
            dp[i][0]=rec[pre[B[i]]];
            int cur=dp[i][0];
            for(int j=1;j<20;j++){
                if(cur!=-1){
                    dp[i][j]=dp[cur][j-1];
                    cur=dp[i][j];
                }else{
                    dp[i][j]=cur;
                }
            }
            int s=n-1;
            int p=i;
            for(int j=19;j>=0&&s>0;j--){
                if((1<<j)<=s){
                    s-=1<<j;
                    p=dp[p][j];
                    if(p==-1){
                        break;
                    }
                }
            }
            d=Math.max(d,p);
            rec2[i]=d;
        }
        StringBuilder sb=new StringBuilder();
        for(int i=0;i<q;i++){
            int l=ni()-1,r=ni()-1;
            if(rec2[r]>=l){
                sb.append(1);
            }else{
                sb.append(0);
            }
        }
        out.println(sb.toString());
    }
    //input
    @SuppressWarnings("unused")
    private ArrayList<Integer>[] ng(int n, int m) {
        ArrayList<Integer>[] graph=(ArrayList<Integer>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            int s=in.nextInt()-1,e=in.nextInt()-1;
            graph[s].add(e);
            graph[e].add(s);
        }
        return graph;
    }

    private ArrayList<long[]>[] ngw(int n, int m) {
        ArrayList<long[]>[] graph=(ArrayList<long[]>[])new ArrayList[n];
        for(int i=0;i<n;i++) {
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++) {
            long s=in.nextLong()-1,e=in.nextLong()-1,w=in.nextLong();
            graph[(int)s].add(new long[] {e,w});
            graph[(int)e].add(new long[] {s,w});
        }
        return graph;
    }

    private int ni() {
        return in.nextInt();
    }

    private long nl() {
        return in.nextLong();
    }
    private double nd() {
        return in.nextDouble();
    }
    private String ns() {
        return in.next();
    }

    private long[] na(int n) {
        long[] A=new long[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextLong();
        }
        return A;
    }

    private int[] nia(int n) {
        int[] A=new int[n];
        for(int i=0;i<n;i++) {
            A[i]=in.nextInt();
        }
        return A;
    }
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;
    InputStreamReader input;//no buffer
    public FastReader()
    {
        br=new BufferedReader(new InputStreamReader(System.in));
    }

    public FastReader(boolean isBuffer)
    {
        if(!isBuffer){
            input=new InputStreamReader(System.in);
        }else{
            br=new BufferedReader(new InputStreamReader(System.in));
        }
    }

    public boolean hasNext(){
        try{
            String s=br.readLine();
            if(s==null){
                return  false;
            }
            st=new StringTokenizer(s);
        }catch(IOException e){
            e.printStackTrace();
        }
        return true;
    }

    public String next()
    {   
        if(input!=null){
            try {
                StringBuilder sb=new StringBuilder();
                int ch=input.read();
                while(ch=='\n'||ch=='\r'||ch==32){
                    ch=input.read();
                }
                while(ch!='\n'&&ch!='\r'&&ch!=32){
                    sb.append((char)ch);
                    ch=input.read();
                }
                return sb.toString();
            }catch (Exception e){
                e.printStackTrace();
            }
        }
        while(st==null || !st.hasMoreElements())//回车，空行情况
        {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }

    public int nextInt()
    {
        return (int)nextLong();
    }

    public long nextLong() {
        try {
            if(input!=null){
                long ret=0;
                int b=input.read();
                while(b<'0'||b>'9'){
                    b=input.read();
                }
                while(b>='0'&&b<='9'){
                    ret=ret*10+b;
                    b=input.read();
                }
                return ret;
            }
        }catch (Exception e){
            e.printStackTrace();
        }
        return Long.parseLong(next());
    }

    public double nextDouble()
    {
        return Double.parseDouble(next());
    }
}