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.
from sys import stdin, stdout

n, m, q = map(int, raw_input().strip().split())
p = map(int, raw_input().strip().split())
a = map(int, raw_input().strip().split())

LOG = 0
k = 1
while k <= n:
    k <<= 1
    LOG += 1

nexx = [0] * (n + 1)
for i in xrange(n):
    if i + 1 < n: nexx[p[i]] = p[i + 1]
    else: nexx[p[i]] = p[0]

jump = [[m for i in xrange(LOG + 1)] for j in xrange(m + 1)]
nearest = [m] * (n + 1)

for i in xrange(m - 1, -1, -1):
    jump[i][0] = nearest[nexx[a[i]]]
    nearest[a[i]] = i

for i in xrange(m - 1, -1, -1):
    for j in xrange(1, LOG + 1):
        jump[i][j] = jump[jump[i][j - 1]][j - 1]

least = [0] * m
for l in xrange(m):
    d = n - 1
    r = l
    for j in xrange(LOG, -1, -1):
        jpow = 1 << j
        if jpow <= d:
            d -= jpow
            r = jump[r][j]
    least[l] = r

INF = int(1e20) # infinitely large number
N = m + 3
t = [0 for i in xrange(2 * N)]

def build():  # build the tree
    for i in xrange(m - 1, 0, -1): t[i] = min(t[i << 1],  t[i << 1 | 1])

def query(l, r):  # range minimum query on interval [l, r) with 0-indexing
    res = INF
    l += m; r += m
    while (l < r):      
        if (l & 1):
            res = min(res, t[l])
            l += 1
        if (r & 1):
            res = min(res, t[r - 1])
            r -= 1
    
        l >>= 1; r >>= 1
    return res

t[m : 2 * m] = least
build()

inp = stdin.readlines()
out = []
for line in inp:
    l, r = map(int, line.strip().split())
    closest = query(l - 1, r)
    if closest < r: out.append('1')
    else: out.append('0')

stdout.write(''.join(out))