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 LynyrdSkynyrd {
    public static void main(String[] args) throws IOException {
        Reader in = new Reader();
        PrintWriter out = new PrintWriter(System.out);
        
        int P = in.nextInt(), N = in.nextInt(), Q = in.nextInt();
        int[] p = new int[P], pos = new int[P];
        for (int i = 0; i < P; i++) {
            p[i] = in.nextInt() - 1;
            pos[p[i]] = i;
        }
        TreeSet<Integer>[] set = new TreeSet[P];
        for (int i = 0; i < P; i++) {
            set[i] = new TreeSet<>();
        }
        int[] arr = new int[N];
        for (int i = 0; i < N; i++) {
            arr[i] = in.nextInt() - 1;
            set[arr[i]].add(i);
        }
        
        int M = Integer.toBinaryString(N).length();
        int[][] jump = new int[N][M];
        for (int i = 0; i < N; i++) {
            int next = p[(pos[arr[i]] + 1) % P];
            jump[i][0] = set[next].isEmpty() || set[next].last() < i ? -1 : set[next].higher(i - 1);
        }
        for (int j = 1; j < M; j++) {
            for (int i = 0; i < N; i++) {
                jump[i][j] = jump[i][j - 1] == -1 ? -1 : jump[jump[i][j - 1]][j - 1];
            }
        }
        
        int[] end = new int[N];
        for (int i = 0; i < N; i++) {
            end[i] = i;
            int n = P - 1;
            for (int j = 0; j < M; j++) {
                if (end[i] == -1) {
                    break;
                }
                if (n % 2 == 1) {
                    end[i] = jump[end[i]][j];
                }
                n >>= 1;
            }
        }
        
        int[] tree = new int[N * 2];
        for (int i = 0; i < N; i++) {
            tree[i + N] = end[i] == -1 ? Integer.MAX_VALUE : end[i];
        }
        for (int i = N - 1; i > 0; i--) {
            tree[i] = Math.min(tree[i * 2], tree[i * 2 + 1]);
        }
        
        char[] res = new char[Q];
        for (int q = 0; q < Q; q++) {
            int l = in.nextInt() - 1 + N, r = in.nextInt() - 1 + N;
            int R = r - N;
            int min = Integer.MAX_VALUE;
            while (l <= r) {
                if (l % 2 == 1) {
                    min = Math.min(min, tree[l]);
                    l >>= 1;
                    l++;
                } else {
                    l >>= 1;
                }
                if (r % 2 == 0) {
                    min = Math.min(min, tree[r]);
                    r >>= 1;
                    r--;
                } else {
                    r >>= 1;
                }
            }
            res[q] = min <= R ? '1' : '0';
        }
        out.println(res);
        
        out.close();
    }
    static class Reader {
        BufferedReader in;
        StringTokenizer st;
        public Reader() {
            in = new BufferedReader(new InputStreamReader(System.in));
            st = new StringTokenizer("");
        }
        public String nextLine() throws IOException {
            st = new StringTokenizer("");
            return in.readLine();
        }
        public String next() throws IOException {
            while (!st.hasMoreTokens()) {
                st = new StringTokenizer(in.readLine());
            }
            return st.nextToken();
        }
        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }
    }
    public static void sort(int[] arr) {
        List<Integer> list = new ArrayList<>();
        for (int i : arr) {
            list.add(i);
        }
        Collections.sort(list);
        for (int i = 0; i < arr.length; i++) {
            arr[i] = list.get(i);
        }
    }
}