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.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;


public class Main {

	static class Task {
		
		int NN = 200005;
		int MOD = 1000000007;
		int INF = 2000000000;
		long INFINITY = 2000000000000000000L;

		int []p;
		int []a;
		int []at;
		int [][]next;
		List<Integer> [] pos;
		int [] optPos;
		
		public void solve(InputReader in, PrintWriter out) {
			int n = in.nextInt(), m = in.nextInt(), q = in.nextInt();
			p = new int[n];
			at = new int[n + 1];
			for(int i=0;i<n;++i) {
				p[i] = in.nextInt();
				at[p[i]] = i;
			}
			a = new int[m];
			for(int i=0;i<m;++i) {
				a[i] = in.nextInt();
			}
			next = new int[m][20];
			pos = new ArrayList[n + 1];
			for(int i=1;i<=n;++i) {
				pos[i] = new ArrayList<>();
			}
			for(int i=0;i<m;++i) {
				pos[a[i]].add(i);
			}
			for(int i=0;i<m;++i) {
				int value = a[i];
				int nextValue = p[(at[value] + 1)%n];
				int lo = 0, hi = pos[nextValue].size() - 1, mid = 0;
				while(lo < hi) {
					mid = (lo + hi) / 2;
					if(pos[nextValue].get(mid) <= i) {
						lo = mid + 1;
					} else {
						hi = mid;
					}
				}
				if(pos[nextValue].isEmpty() || pos[nextValue].get(lo) <= i) {
					next[i][0] = -1;
				} else {
					next[i][0] =pos[nextValue].get(lo); 
				}
			}
			for(int j=1;j<20;++j) {
				for(int i=0;i<m;++i) {
					int nextPos = next[i][j - 1];
					if(nextPos != -1) {
						next[i][j] = next[nextPos][j - 1];
					} else {
						next[i][j] = -1;
					}
				}
			}
			optPos = new int[m];
			for(int i=0;i<m;++i) {
				int tot = n - 1;
				int l = i;
				for(int j=19;j>=0;--j) {
					if((1<<j)<=tot && l != -1) {
						tot -= (1<<j);
						l = next[l][j];
					}
				}
				if(l == -1)
					l = INF;
				optPos[i] = l;
			}
			T = new int[5*m];
			build(1, 0, m - 1);
			while(q-->0) {
				int l = in.nextInt(), r = in.nextInt();
				--l;--r;
				int optimalPostion = query(1, 0, m - 1, l, r);
				out.print(optimalPostion <= r ? 1 : 0);
			}
			out.println("");
		}
		
		int query(int i, int l, int r, int x, int y) {
			if(x > r || y < l)
				return INF;
			if(x <= l && r <= y)
				return T[i];
			int m = (l + r) / 2;
			return Math.min(
					query(2*i, l, m, x, y),
					query(1+2*i, 1+m, r, x, y));
		}
		
		void build(int i, int l, int r) {
			if(l == r) {
				T[i] = optPos[l];return;
			}
			int m = (l + r) / 2;
			build(i*2, l, m);
			build(1+2*i, 1+m, r);
			T[i] = Math.min(T[2*i], T[1+2*i]);
		}
		
		int [] T;
		
	}
	
	static void prepareIO(boolean isFileIO) {
		//long t1 = System.currentTimeMillis();
		Task solver = new Task();
		// Standard IO
		if(!isFileIO) { 
			InputStream inputStream = System.in;
	        OutputStream outputStream = System.out;
	        InputReader in = new InputReader(inputStream);
	        PrintWriter out = new PrintWriter(outputStream);
	        solver.solve(in, out);
	        //out.println("time(s): " + (1.0*(System.currentTimeMillis()-t1))/1000.0);
	        out.close();
		}
        // File IO
		else {
			String IPfilePath = System.getProperty("user.home") + "/Downloads/ip.in";
	        String OPfilePath = System.getProperty("user.home") + "/Downloads/op.out";
	        InputReader fin = new InputReader(IPfilePath);
	        PrintWriter fout = null;
	        try {
				fout = new PrintWriter(new File(OPfilePath));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
	        solver.solve(fin, fout);
	        //fout.println("time(s): " + (1.0*(System.currentTimeMillis()-t1))/1000.0);
	        fout.close();
		}
	}
	
	public static void main(String[] args) {
        prepareIO(false);
	}
	
	static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }
        
        public InputReader(String filePath) {
        	File file = new File(filePath);
            try {
				reader = new BufferedReader(new FileReader(file));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
            tokenizer = null;
        }
        
        public String nextLine() {
        	String str = "";
        	try {
				str = reader.readLine();
			} catch (IOException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
        	return str;
        }
        
        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());
        }

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

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

}