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
The above is tricky. Write me a correct solution in PYTHON3.
assistant
#Here is the Python code you requested.

mod = 1000000007
eps = 10**-9

def main():
    import sys
    input = sys.stdin.readline

    from collections import deque

    # compressed[v]: 縮約後のグラフで縮約前のvが属する頂点
    # num: 縮約後のグラフの頂点数
    def SCC(adj, adj_rev):
        N = len(adj) - 1
        seen = [0] * (N + 1)
        compressed = [0] * (N + 1)
        order = []

        for v0 in range(1, N + 1):
            if seen[v0]:
                continue
            st = deque()
            st.append(v0)
            while st:
                v = st.pop()
                if v < 0:
                    order.append(-v)
                else:
                    if seen[v]:
                        continue
                    seen[v] = 1
                    st.append(-v)
                    for u in adj[v]:
                        st.append(u)

        #print(order)
        seen = [0] * (N + 1)
        num = 0
        for v0 in reversed(order):
            if seen[v0]:
                continue
            num += 1
            st = deque()
            st.append(v0)
            seen[v0] = 1
            compressed[v0] = num
            #print(v0, num)
            while st:
                v = st.pop()
                for u in adj_rev[v]:
                    if seen[u]:
                        continue
                    #print([v, u])
                    seen[u] = 1
                    compressed[u] = num
                    st.append(u)

        return num, compressed

    # 縮約後のグラフを構築
    # 先にSCC()を実行してnum, compressedを作っておく
    def construct(adj, num, compressed):
        N = len(adj) - 1
        adj_compressed = [set() for _ in range(num + 1)]
        for v in range(1, N + 1):
            v_cmp = compressed[v]
            for u in adj[v]:
                u_cmp = compressed[u]
                if v_cmp != u_cmp:
                    adj_compressed[v_cmp].add(u_cmp)
        return adj_compressed

    N, M = map(int, input().split())
    adj = [[] for _ in range(N+1)]
    adj_rev = [[] for _ in range(N+1)]
    for _ in range(M):
        a, b = map(int,...