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.*;
import java.util.*;
  
public class Main {
    public static void main(String... args) { new Main().work(); }
    
    void work() { WORK(); out.close(); }
    
    class InputReader {
    	BufferedReader br;
    	StringTokenizer st;
    	
    	InputReader(InputStream stream) {
    		br = new BufferedReader(new InputStreamReader(stream));
    		st = null;
		}
    	
    	String next() {
    		while(st==null || !st.hasMoreTokens()) {
    			try { st = new StringTokenizer(br.readLine()); }
    			catch(IOException e) { e.printStackTrace(); }
    		}
    		return st.nextToken();
    	}
    	
    	int nextInt() { return Integer.parseInt(next()); }
    	long nextLong() { return Long.parseLong(next()); }
    }
    
    InputReader in = new InputReader(System.in);
    PrintWriter out = new PrintWriter(System.out);
    
    final int N = (int)(2e5+10), MOD = (int)(1e9+7);
    
    int a[] = new int[N], b[] = new int[N], now[] = new int[N], dp[][] = new int[N][30];
    int Log[] = new int[N], p[][] = new int[N][30];
    
    void WORK() {
    	int n = in.nextInt(), m = in.nextInt(), q = in.nextInt();
    	for(int i = 1; i <= n; i++) a[in.nextInt()] = i;
    	Log[0] = -1;
    	for(int i = 1; i <= m; i++) {
    		b[i] = in.nextInt();
    		Log[i] = Log[i>>1]+1;
    	}
    	for(int i = m; i >= 1; i--) {
    		now[a[b[i]]] = i;
    		int t = a[b[i]]==n?1:a[b[i]]+1;
    		if(now[t] != 0) p[i][0] = now[t];
    	}
    	
    	for(int i = 1; i <= 20; i++)
    		for(int j = 1; j <= m; j++)
    			p[j][i] = p[p[j][i-1]][i-1];
    	for(int i = 1; i <= m; i++) {
//    		out.println("! = " + p[i][0]);
    		dp[i][0] = i;
    		int z = n-1;
    		for(int j = 20; j >= 0; j--) if(z >= (1<<j)) {
    			z -= (1<<j);
    			dp[i][0] = p[dp[i][0]][j];
    		}
    		if(dp[i][0] == 0) dp[i][0] = m+1;
//    		out.println("? = " + dp[i][0]);
    	}
    	for(int j = 1; j <= 20; j++) {
    		for(int i = 1; i+(1<<j)-1 <= m; i++) {
    			dp[i][j] = Math.min(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    		}
    	}
    	while(q-- > 0) {
    		int l = in.nextInt(), r = in.nextInt();
    		int k = Log[r-l+1];
    		int t = Math.min(dp[l][k], dp[r-(1<<k)+1][k]);
//    		out.println("t = " + dp[l][k] + "k = " + dp[r-(1<<k)+1][k]);
    		if(Math.min(dp[l][k], dp[r-(1<<k)+1][k]) <= r) out.print(1);
    		else out.print(0);
    	}
    }
}
/*
10 100 1
4 7 1 2 3 8 6 9 10 5
7 6 4 9 3 10 2 8 5 4 7 5 9 1 5 10 10 2 4 6 8 4 6 4 7 5 7 7 6 5 10 7 8 8 10 5 7 3 10 9 5 2 3 9 6 4 4 5 6 1 6 1 9 10 2 9 2 8 4 10 4 8 9 4 6 2 9 10 9 1 6 10 1 9 1 4 9 5 5 10 4 5 3 1 8 7 7 2 1 7 5 5 10 7 5 9 9 2 8 3
27 76
*/