1142_B. Lynyrd Skynyrd
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.
```python
import os
from io import BytesIO, StringIO
input = BytesIO(os.read(0, os.fstat(0).st_size)).readline

def input_as_list():
    return list(map(int, input().split()))

def array_of(f, *dim):
    return [array_of(f, *dim[1:]) for _ in range(dim[0])] if dim else f()

def main():
    n, m, q = input_as_list()
    p = input_as_list()
    a = input_as_list()

    if n == 1:
        print('1' * q)
        return

    next_p = array_of(int, n)
    next_a = array_of(int, m)

    prev = p[-1] - 1
    for pi in p:
        pi -= 1
        next_p[prev] = pi
        prev = pi

    idx_a = array_of(int, n)
    for i in range(len(a) - 1, -1, -1):
        ai = a[i] - 1
        next_a[i] = idx_a[next_p[ai]]
        idx_a[ai] = i


    ep = array_of(int, m)
    parent = next_a
    children = array_of(list, m)
    roots = []
    for i, p in enumerate(parent):
        if p == 0:
            roots.append(i)
        else:
            children[p].append(i)

    rank = array_of(int, m)
    parent_ex = array_of(list, m)
    for r in roots:
        stack = [r]
        while stack:
            p = stack.pop()
            for c in children[p]:
                rank[c] = rank[p] + 1
                parent_ex[c].append(p)
                i = 1
                while 2 ** i <= rank[c]:
                    parent_ex[c].append(parent_ex[parent_ex[c][-1]][i - 1])
                    i += 1
                stack.append(c)

    for i in range(m):
        y = i
        s = 0
        try:
            for j in reversed(range(20)):
                if n - s - 1 >= 2**j:
                    y = parent_ex[y][j]
                    s += 2**j
            ep[i] = y
        except IndexError:
            pass

    mn = 200000
    flag = False
    for i in range(len(ep) - 1, -1, -1):
        epi = ep[i]
        if epi != 0:
            if epi > mn:
                ep[i] = mn
            mn = min(mn, epi)
            flag = True
        else:
            if flag:
                ep[i] = mn

    out = []
    for _ in range(q):
        l, r = map(int, input().split())
        l, r = l - 1, r - 1
        if ep[l] == 0 or ep[l] > r:
            out.append('0')
        else:
            out.append('1')

    print(''.join(out))
    
main()
```