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 B1142 {
	static int n, p[], idxInP[];

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		n = sc.nextInt();
		int m = sc.nextInt();
		int q = sc.nextInt();
		p = sc.nextIntArr(n);
		idxInP = new int[n];
		for (int i = 0; i < p.length; i++) {
			idxInP[--p[i]] = i;
		}
		int[] arr = sc.nextIntArr(m);
		int[] lastOcc = new int[n];
		Arrays.fill(lastOcc, -1);
		int[][] par = new int[20][m];
		for (int i = 0; i < m; i++) {
			int cur = --arr[i];
			int prevValInP = (idxInP[cur] - 1);
			if (prevValInP == -1)
				prevValInP += n;
			prevValInP = p[prevValInP];
			par[0][i] = lastOcc[prevValInP];
			lastOcc[cur] = i;
		}
		for (int lvl = 1; lvl < 20; lvl++) {
			for (int i = 0; i < m; i++) {
				par[lvl][i] = par[lvl - 1][i] == -1 ? -1 : par[lvl - 1][par[lvl - 1][i]];
			}
		}
		int[] lim = new int[m];
		for (int i = 0; i < m; i++) {
			int anscestor = n - 1;
			int cur = i;
			int lvl = 0;
			while (anscestor != 0) {
				if ((anscestor & 1) == 1) {
					cur = cur == -1 ? -1 : par[lvl][cur];
				}
				lvl++;
				anscestor >>= 1;
			}
			lim[i] = Math.max(i == 0 ? -1 : lim[i - 1], cur);
		}
//		pw.println(Arrays.toString(lim));
		while (q-- > 0) {
			int l = sc.nextInt() - 1;
			int r = sc.nextInt() - 1;
			pw.print(l <= lim[r] ? 1 : 0);
		}
		pw.println();
		pw.close();
	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

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

		public Scanner(FileReader f) {
			br = new BufferedReader(f);
		}

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

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = Integer.parseInt(next());
			}
			return arr;
		}

	}

}
