human
A direced graph is strongly connected if every two nodes are reachable from each other. In a strongly connected component of a directed graph, every two nodes of the component are mutually reachable.

Constraints

* 1 ≤ |V| ≤ 10,000
* 0 ≤ |E| ≤ 30,000
* 1 ≤ Q ≤ 100,000

Input

A directed graph G(V, E) and a sequence of queries where each query contains a pair of nodes u and v.

|V| |E|
s0 t0
s1 t1
:
s|E|-1 t|E|-1
Q
u0 v0
u1 v1
:
uQ-1 vQ-1

|V| is the number of nodes and |E| is the number of edges in the graph. The graph nodes are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target nodes of i-th edge (directed).

ui and vi represent a pair of nodes given as the i-th query.

Output

For each query, pinrt "1" if the given nodes belong to the same strongly connected component, "0" otherwise.

Example

Input

5 6
0 1
1 0
1 2
2 4
4 3
3 2
4
0 1
0 3
2 3
3 4

Output

1
0
1
1
 ... Give me a the Python program you requested?
assistant
#Here is the Python program you requested.

import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
import time,random

sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)

def scc(n, e, re):

    v = [None] * n
    t = []
    def dfs(i):
        v[i] = 0
        for b in e[i]:
            if v[b] is None:
                dfs(b)
        t.append(i)

    for i in range(n):
        if v[i] is None:
            dfs(i)

    r = [None] * n
    def dfs2(i, p):
        r[i] = p
        for b in re[i]:
            if r[b] is None:
                dfs2(b, p)

    for c in t[::-1]:
        if r[c] is None:
            dfs2(c, c)

    return r

def main():
    n,m = LI()
    st = [LI() for _ in range(m)]
    q = I()
    uv = [LI() for _ in range(q)]

    e = collections.defaultdict(set)
    re = collections.defaultdict(set)
    for s,t in st:
        e[s].add(t)
        re[t].add(s)

    r = scc(n, e, re)
    rr = []
    for u,v in uv:
        if r[u] == r[v]:
            rr.append(1)
        else:
            rr.append(0)

    return JA(rr, "\n")

# start = time.time()
print(main())
# pe(time.time() - start)