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 sys
 
class segmentTree:
	def __init__(self, n):
		self.n = n
		self.seg = [self.n + 1] * (self.n << 1)
		
	def update(self, p, value):
		p += self.n
		self.seg[p] = value
		while p > 1:
			p >>= 1
			self.seg[p] = min(self.seg[p * 2], self.seg[p * 2 + 1])
			
	
	def query(self, l, r):
		res = self.n
		l += self.n
		r += self.n
		while l < r:
			if l & 1:
				res = min(res, self.seg[l])
				l += 1
			if r & 1:
				res = min(res, self.seg[r - 1])
				r -= 1
			l >>= 1
			r >>= 1
		return res
				
inp = [int(x) for x in sys.stdin.read().split()]
 
n, m, q = inp[0], inp[1], inp[2]
 
p = [inp[idx] for idx in range(3, n + 3)]
 
index_arr = [0] * (n + 1)
for i in range(n):	index_arr[p[i]] = i
 
a = [inp[idx] for idx in range(n + 3, n + 3 + m)]
 
leftmost_pos = [m] * (n + 1)
next = [-1] * m
 
for i in range(m - 1, -1, -1):
	index = index_arr[a[i]]
	right_index = 0 if index == n - 1 else index + 1
	right = p[right_index]
	next[i] = leftmost_pos[right]
	leftmost_pos[a[i]] = i
	
log = 0
while (1 << log) <= n: log += 1
log += 1
dp = [[m for _ in range(m + 1)] for  _ in range(log)]
 
for i in range(m):
	dp[0][i] = next[i]
 
for j in range(1, log):
	for i in range(m):
		dp[j][i] = dp[j - 1][dp[j - 1][i]]
 
tree = segmentTree(m)
for i in range(m):
	p = i
	len = n - 1
	for j in range(log - 1, -1, -1):
		if (1 << j) <= len:
			p = dp[j][p]
			len -= (1 << j)
	tree.update(i, p)
 
inp_idx = n + m + 3
ans = []
for i in range(q):
	l, r = inp[inp_idx] - 1, inp[inp_idx + 1] - 1
	inp_idx += 2
	if tree.query(l, r + 1) <= r:
		ans.append('1')
	else:
		ans.append('0')
print(''.join(ans))