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.*;
import java.math.*;

public class B {
	static byte[] buf = new byte[1<<26];
    static int bp = -1;	
	
	public static void main(String[] args) throws IOException {

		/**/
		DataInputStream in = new DataInputStream(System.in);
		/*/
		DataInputStream in = new DataInputStream(new FileInputStream("src/b.in"));
		/**/
		
		in.read(buf, 0, 1<<26);
		int n = nni();
		int m = nni();
		int q = nni();
		
		int[] p = new int[n];
		int[] a = new int[m];
		for (int i = 0; i < n; i++)
			p[i] = nni();
		for (int i = 0; i < m; i++)
			a[i] = nni();
		
		int[] np = new int[n+1];
		np[p[n-1]] = p[0];
		for (int i = 0; i < n-1; i++)
			np[p[i]] = p[i+1];
		
		HashMap<Integer, TreeSet<Integer>> tm = new HashMap<>();
		for (int i = 0; i <= n; i++) {
			tm.put(i, new TreeSet<>());
		}
		for (int i = 0; i < m; i++) {
			if (!tm.containsKey(a[i]))
				tm.put(a[i], new TreeSet<>());
			tm.get(a[i]).add(i);
		}
		for (int key : tm.keySet()) {
			tm.get(key).add(m);
		}
		int[] next = new int[m+1];
		for (int i = 0; i < m; i++) {
			next[i] = tm.get(np[a[i]]).higher(i);
		}
		next[m] = m;
		int[][] pn = new int[20][m+1];
		for (int i = 0; i <= m; i++)
			pn[0][i] = next[i];
		for (int i = 1; i < 20; i++) {
			for (int j = 0; j <= m; ++j)
				pn[i][j] = pn[i-1][pn[i-1][j]];
		}
		int[] ce = new int[m];
		int jumps = n-1;
		for (int i = 0; i < m; i++) {
			int curr = i;
			for (int j = 19; j >= 0; --j) {
				if ((jumps&(1<<j))>0) {
					curr = pn[j][curr];
					if (curr==m)
						break;
				}
			}
			ce[i] = curr;
		}
		int[] cem = new int[m];
		cem[m-1] = ce[m-1];
		for (int i = m-2; i >= 0; --i) {
			cem[i] = Math.min(cem[i+1], ce[i]);
		}
		
		StringBuilder ans = new StringBuilder();
		for (int i = 0; i < q; i++) {
			int l = nni()-1;
			int r = nni()-1;
			if (cem[l] <= r)
				ans.append("1");
			else
				ans.append("0");
		}
		System.out.println(ans);
	}
	
	public static int nni() {
        int ret = 0;
        byte b = buf[++bp];
        while (true) {
            ret = ret*10+b-'0';
            b = buf[++bp];
            if (b<'0'||b>'9') {
            	while (buf[bp+1]=='\r'||buf[bp+1]=='\n'||buf[bp+1]==' ') {++bp;}
            	break;
            }
        }
        return ret;
    }
}