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.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
import java.util.TreeSet;

@SuppressWarnings("unchecked")
public class E1143_2 {
	
	int n, m, q, logm;
	int[] perm, index, a, vals;
	TreeSet<Integer>[] occ;
	ArrayList<Integer>[] vs;
	int[][] par;
	
	void buildGraph() {
		vs = new ArrayList[m];
		int[] indeg = new int[m];
		for(int i = 0; i < m; ++i) vs[i] = new ArrayList<>();
		for(int i = 0; i < m; ++i) {
			int x = a[i];
			int p = index[x];
			int prev = (p - 1 + n) % n;
			int pEle = perm[prev];
			Integer j = occ[pEle].lower(i);
			if(j == null) continue;
			vs[j].add(i);
			++indeg[i];
		}
		logm = Integer.numberOfTrailingZeros(Integer.highestOneBit(m)) + 1;
		par = new int[logm][m];
		for(int i = 0; i < logm; ++i) Arrays.fill(par[i], -1);
		ArrayDeque<Integer> q = new ArrayDeque<>();
		for(int i = 0; i < m; ++i) if(indeg[i] == 0) q.add(i);
		while(!q.isEmpty()) {
			int node = q.poll();
			for(int i : vs[node]) {
				par[0][i] = node;
				q.add(i);
			}
		}
		for(int log = 0; log < logm - 1; ++log) {
			for(int i = 0; i < m; ++i) {
				if(par[log][i] != -1) {
					par[log + 1][i] = par[log][par[log][i]];
				}
			}
		}
	}
	
	int walk(int node, int dist) {
		for(int log = logm - 1; log >= 0; --log) {
			int d = 1 << log;
			if(d > dist) continue;
			dist -= d;
			node = par[log][node];
			if(node == -1) return -1;
		}
		return node;
	}
	
	public void solve(Scanner in, PrintWriter out) {
		n = in.nextInt();
		m = in.nextInt();
		q = in.nextInt();
		a = new int[m];
		perm = new int[n];
		index = new int[n];
		vals = new int[m];
		occ = new TreeSet[n];
		Arrays.fill(vals, -1);
		for(int i = 0; i < n; ++i) occ[i] = new TreeSet<>();
		for(int i = 0; i < n; ++i) perm[i] = in.nextInt() - 1;
		for(int i = 0; i < m; ++i) a[i] = in.nextInt() - 1;
		for(int i = 0; i < m; ++i) occ[a[i]].add(i);
		for(int i = 0; i < n; ++i) index[perm[i]] = i;
		buildGraph();
		for(int i = 0; i < m; ++i) {
			int prev = walk(i, n - 1);
			vals[i] = prev;
		}
		RMQ rmq = new RMQ(vals);
		while(q-->0) {
			int l = in.nextInt() - 1;
			int r = in.nextInt() - 1;
			int maxL = rmq.query(l, r);
			out.print(maxL >= l ? 1 : 0);
		}
	}
	
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		new E1143_2().solve(in, out);
		in.close();
		out.close();
	}
	
	static class RMQ {
		int[] vs;
		int[][] lift;

		public RMQ(int[] vs) {
			this.vs = vs;
			int n = vs.length;
			int maxlog = Integer.numberOfTrailingZeros(Integer.highestOneBit(n)) + 2;
			lift = new int[maxlog][n];
			for (int i = 0; i < n; i++)
				lift[0][i] = vs[i];
			int lastRange = 1;
			for (int lg = 1; lg < maxlog; lg++) {
				for (int i = 0; i < n; i++) {
					lift[lg][i] = Math.max(lift[lg - 1][i], lift[lg - 1][Math.min(i + lastRange, n - 1)]);
				}
				lastRange *= 2;
			}
		}

		public int query(int low, int hi) {
			int range = hi - low + 1;
			int exp = Integer.highestOneBit(range);
			int lg = Integer.numberOfTrailingZeros(exp);
			return Math.max(lift[lg][low], lift[lg][hi - exp + 1]);
		}
	}
}
