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.
//package com.company;
// Always comment out package when submitting.

import java.io.*;
import java.util.*;

public class Main {
    public static class Task {

        public void solve(Scanner sc, PrintWriter pw) throws IOException {
            int n = sc.nextInt();
            int m = sc.nextInt();
            int q = sc.nextInt();
            int[] p = new int[n];
            for (int i = 0; i < n; i++) {
                p[i] = sc.nextInt() - 1;
            }
            int[] a = new int[m];
            for (int i = 0; i < m; i++) {
                a[i] = sc.nextInt() - 1;
            }
            int[] next = new int[m];
            Arrays.fill(next, -1);
            List<Integer>[] vals = new List[n];
            for (int i = 0; i < n; i++) {
                vals[i] = new ArrayList<>();
            }
            for (int i = 0; i < m; i++) {
                vals[a[i]].add(i);
            }
            for (int i = 0; i < n; i++) {
                int cur = p[i];
                int nex = p[(i + 1) % n];
                List<Integer> i1 = vals[cur];
                List<Integer> i2 = vals[nex];
                int idx1 = 0, idx2 = 0;
                while (idx1 < i1.size() && idx2 < i2.size()) {
                    if (i1.get(idx1) < i2.get(idx2)) {
                        next[i1.get(idx1)] = i2.get(idx2);
                        idx1++;
                    } else {
                        idx2++;
                    }
                }
            }
            int[][] jumpTable = new int[m][20];
            for (int i = 0; i < m; i++) {
                jumpTable[i][0] = next[i];
            }
            for (int i = 1; i < 20; i++) {
                for (int j = 0; j < m; j++) {
                    if (jumpTable[j][i - 1] == -1) {
                        jumpTable[j][i] = -1;
                    } else {
                        jumpTable[j][i] = jumpTable[jumpTable[j][i - 1]][i - 1];
                    }
                }
            }
            StringBuilder sb = new StringBuilder();
            int[] reach = new int[m];
            for (int i = 0; i < m; i++) {
                int mv = i;
                int count = 0;
                for (int j = 19; j >= 0; j--) {
                    if ((1 << j) + count <= n - 1) {
                        mv = jumpTable[mv][j];
                        if (mv == -1) {
                            break;
                        }
                        count += 1 << j;
                    }
                }
                reach[i] = mv;
                if (reach[i] == -1) reach[i] = m;
            }
            int[][] rangeMin = new int[m][20];
            for (int i = 0; i < m; i++) {
                rangeMin[i][0] = reach[i];
            }
            for (int i = 1; i < 20; i++) {
                for (int j = 0; j + (1 << (i - 1)) < m; j++) {
                    rangeMin[j][i] = Math.min(rangeMin[j][i - 1], rangeMin[j + (1 << (i - 1))][i - 1]);
                }
            }
            for (int i = 0; i < q; i++) {
                int l = sc.nextInt() - 1;
                int r = sc.nextInt();
                int count = m;
                int mv = l;
                for (int j = 19; j >= 0; j--) {
                    if (mv + (1 << j) <= r) {
                        count = Math.min(count, rangeMin[mv][j]);
                        mv += 1 << j;
                    }
                }
                if (count < r) {
                    sb.append(1);
                } else {
                    sb.append(0);
                }
            }
            pw.println(sb.toString());
        }
    }

    // template, actual code is in class Task.
    static long TIME_START, TIME_END;

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
//        Scanner sc = new Scanner(new FileInputStream("Test.in"));
        PrintWriter pw = new PrintWriter(new BufferedOutputStream(System.out));
//        PrintWriter pw = new PrintWriter(new FileOutputStream("Test.out"));
        Runtime runtime = Runtime.getRuntime();
        long usedMemoryBefore = runtime.totalMemory() - runtime.freeMemory();
        TIME_START = System.currentTimeMillis();
        Task t = new Task();
        t.solve(sc, pw);
        TIME_END = System.currentTimeMillis();
        long usedMemoryAfter = runtime.totalMemory() - runtime.freeMemory();
        pw.close();
//        System.err.println("Memory increased: " + (usedMemoryAfter - usedMemoryBefore) / 1000000);
//        System.err.println("Time used: " + (TIME_END - TIME_START) + ".");
    }

    // Faster IO with BufferedReader wrapped with Scanner
    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(FileReader s) throws FileNotFoundException {
            br = new BufferedReader(s);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

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

        public boolean ready() throws IOException {
            return br.ready();
        }
    }
}