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

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BLynyrdSkynyrd solver = new BLynyrdSkynyrd();
        solver.solve(1, in, out);
        out.close();
    }

    static class BLynyrdSkynyrd {

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            int q = in.nextInt();
            int[] p = new int[n];
            int[] a = new int[m];

            for (int i = 0; i < n; i++) {
                p[i] = in.nextInt() - 1;
            }
            int[] mp = new int[n];
            for (int i = 0; i < n - 1; i++) {
                mp[p[i]] = p[i + 1];
            }

            mp[p[n - 1]] = p[0];

            for (int i = 0; i < m; i++) {
                a[i] = in.nextInt() - 1;
            }

            int[] last = new int[n];
            Arrays.fill(last, -1);
            BinaryLifting.Vertex[] vs = BinaryLifting.Vertex.emptyGraph(m + 1);

            for (int i = m - 1; i >= 0; i--) {
                int next = mp[a[i]];
                if (last[next] != -1) {
                    vs[last[next]].adj.add(vs[i]);
                } else {
                    vs[m].adj.add(vs[i]);
                }
                last[a[i]] = i;
            }

            BinaryLifting binLift = new BinaryLifting(vs, vs[m]);

            int[] min = new int[m];
            for (int i = m - 1; i >= 0; i--) {
                BinaryLifting.Vertex parent = binLift.ithParent(vs[i], n - 1);
                if (i == m - 1) {
                    min[i] = parent.idx;
                } else {
                    min[i] = Math.min(parent.idx, min[i + 1]);
                }
            }

            for (int i = 0; i < q; i++) {
                int l = in.nextInt() - 1;
                int r = in.nextInt() - 1;
                if (min[l] <= r) {
                    out.print('1');
                } else {
                    out.print('0');
                }
            }
        }
    }

    static class InputReader {

        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
    }

    static class BinaryLifting {

        BinaryLifting.Vertex[] vs;
        int timer;

        public BinaryLifting(BinaryLifting.Vertex[] vs, BinaryLifting.Vertex head) {
            this.vs = vs;
            this.timer = 1;
            dfs(head, head);
        }

        public void dfs(BinaryLifting.Vertex v, BinaryLifting.Vertex p) {
            v.tin = ++timer;
            v.up[0] = p;
            for (int i = 1; i < v.up.length; ++i) {
                v.up[i] = v.up[i - 1].up[i - 1];
            }
            for (BinaryLifting.Vertex to : v.adj) {
                //noinspection ObjectEquality
                if (to != p) {
                    dfs(to, v);
                }
            }
            v.tout = ++timer;
        }

        public BinaryLifting.Vertex ithParent(BinaryLifting.Vertex a, int i) {
            while (i != 0) {
                int pow = a.up.length - 1;
                while ((1 << pow) > i) {
                    pow--;
                }
                a = a.up[pow];
                i -= (1 << pow);
            }
            return a;
        }

        public static class Vertex {

            int idx;
            int tin;
            int tout;
            List<BinaryLifting.Vertex> adj = new ArrayList<>();
            BinaryLifting.Vertex[] up;

            public static BinaryLifting.Vertex[] emptyGraph(int n) {
                int l = 1;
                while ((1 << l) <= n) {
                    ++l;
                }
                BinaryLifting.Vertex[] vs = new BinaryLifting.Vertex[n];
                for (int i = 0; i < n; i++) {
                    vs[i] = new BinaryLifting.Vertex();
                    vs[i].idx = i;
                    vs[i].up = new BinaryLifting.Vertex[l + 1];
                }
                return vs;
            }
        }
    }
}

