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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;


public class B1142 {
    static boolean[] vis;
    static ArrayList<Integer>[] adj;
    
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(System.out);

        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        int q = Integer.parseInt(st.nextToken());
        
        int[] locs = new int[n+1];
        int[] last = new int[n+1];
        int[] ret = new int[m];
        adj = new ArrayList[m];
        for (int i = 0; i < adj.length; i++) {
            adj[i] = new ArrayList<>();
        }
        st = new StringTokenizer(br.readLine());
        for(int i = 0; i<n; i++){
            int c = Integer.parseInt(st.nextToken());
            locs[c] = i;
            last[i] = -1;
        }
        st = new StringTokenizer(br.readLine());
        for(int i = 0; i<m; i++){
            int u = Integer.parseInt(st.nextToken());
            if(u < 0 || u > n){
                ret[i] = -1;
                continue;
            }
            last[locs[u]] = i;
            int bac = last[(locs[u]-1 + n)%n];
            if(bac != -1){
                adj[bac].add(i);
            }
        }
        vis = new boolean[m];
        for(int i = 0; i<m; i++){
            if(!vis[i]){
                dfs(i, new ArrayList(), ret, n);
            }
        }
//        System.out.println(Arrays.toString(ret));
        int[][] qs = new int[q][3];
        for(int i = 0; i<q; i++){
            st = new StringTokenizer(br.readLine());
            qs[i] = new int[]{
                Integer.parseInt(st.nextToken())-1,
                Integer.parseInt(st.nextToken())-1,
                i
            };
        }
        
        Arrays.sort(qs, (int[] a, int[] b) -> a[1] - b[1]);
        boolean[] ans = new boolean[q];
        int max = -1;
        int p = 0;
        for(int i = 0; i<q; i++){
            while(p <= qs[i][1]){
                max = Math.max(max, ret[p]);
                p++;
            }
            ans[qs[i][2]] = qs[i][0] <= max;
        }
        
        for(boolean i : ans){
            pw.print(i ? 1 : 0);
        }
        pw.close();
        br.close();
    }
    
    static void dfs(int u, ArrayList<Integer> pref, int[] ret, int size){
        pref.add(u);
//        System.out.println(u + " " + pref.size());
        if(pref.size() >= size){
            ret[u] = pref.get(pref.size() - size);
        }else{
            ret[u] = -1;
        }
        for(int n : adj[u]){
            if(n != u) dfs(n, pref, ret, size);
        }
        vis[u] = true;
        pref.remove(pref.size()-1);
    }
}
