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 javax.management.loading.MLet;
import java.util.*;
import java.io.*;

public class LynyrdSkynyrd2 {
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int N = scanner.nextInt();
        int M = scanner.nextInt();
        int Q = scanner.nextInt();
        int[] perm = new int[N];
        int[] invPerm = new int[N+1];
        for(int i = 0; i < N; i++) {
            perm[i] = scanner.nextInt();
            invPerm[perm[i]-1] = i;
        }
        int[] prev = new int[N+1];
        int[] parent = new int[M+1];
        Arrays.fill(parent, -1);
        Arrays.fill(prev, -1);
        int[] arr = new int[M+1];
        for(int i = 1; i <= M; i++) arr[i] = scanner.nextInt();
        for(int i = 1; i <= M; i++) {
            parent[i] = prev[perm[(invPerm[arr[i]-1]-1+N)%N]-1];
            prev[arr[i]-1] = i;
        }
        //size for binary lifting table
        int sz = (int) (Math.log(M)/Math.log(2));
        int[][] bl = new int[M+1][sz+1];
        for(int[] dd: bl) Arrays.fill(dd, -1);
        for(int i = 1; i <= M; i++) {
            bl[i][0] = parent[i];
        }
        //update table
        for(int i = 1; i <= sz; i++) {
            for(int j = 1; j <= M; j++) {
                if (bl[j][i-1] != -1) bl[j][i] = bl[bl[j][i-1]][i-1];
            }
        }
        int[] pref = new int[M+1];
        pref[0] = -1;
        for(int i = 1; i <= M; i++) {
//            pref[i] = Math.min(pref[i-1], bl[i][sz]);
            int cur = i;
            for(int d = sz; d >= 0; d--) {
                if (((N-1) & (1 << d)) > 0) cur = bl[cur][d];
                if (cur == -1) break;
            }
            pref[i] = Math.max(pref[i-1],cur);
        }
        while(Q-->0) {
            int l = scanner.nextInt();
            int r = scanner.nextInt();
            if (pref[r] >= l && pref[r] <= r) out.print(1);
            else out.print(0);
        }
        out.println();
        out.flush();
    }
    
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        
        public FastScanner() {
            this(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
            return Long.parseLong(next());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
        
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        
        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextInt();
            }
            return a;
        }
    }
}
