Ken loves ken-ken-pa (Japanese version of hopscotch). Today, he will play it on a directed graph G. G consists of N vertices numbered 1 to N, and M edges. The i-th edge points from Vertex u_i to Vertex v_i.

First, Ken stands on Vertex S. He wants to reach Vertex T by repeating ken-ken-pa. In one ken-ken-pa, he does the following exactly three times: follow an edge pointing from the vertex on which he is standing.

Determine if he can reach Vertex T by repeating ken-ken-pa. If the answer is yes, find the minimum number of ken-ken-pa needed to reach Vertex T. Note that visiting Vertex T in the middle of a ken-ken-pa does not count as reaching Vertex T by repeating ken-ken-pa.

Constraints

* 2 \leq N \leq 10^5
* 0 \leq M \leq \min(10^5, N (N-1))
* 1 \leq u_i, v_i \leq N(1 \leq i \leq M)
* u_i \neq v_i (1 \leq i \leq M)
* If i \neq j, (u_i, v_i) \neq (u_j, v_j).
* 1 \leq S, T \leq N
* S \neq T

Input

Input is given from Standard Input in the following format:


N M
u_1 v_1
:
u_M v_M
S T


Output

If Ken cannot reach Vertex T from Vertex S by repeating ken-ken-pa, print -1. If he can, print the minimum number of ken-ken-pa needed to reach vertex T.

Examples

Input

4 4
1 2
2 3
3 4
4 1
1 3


Output

2


Input

3 3
1 2
2 3
3 1
1 2


Output

-1


Input

2 0
1 2


Output

-1


Input

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6


Output

2
#include <iostream>
#include <queue>
#include <utility>
#include <map>
#include <algorithm>
using namespace std;

typedef long long ll;

int n, m, s, t, a, b, v[100005][3];
vector<int> g[100005];

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 0; i < m; ++i)
        cin >> a >> b, g[a].push_back(b);
    cin >> s >> t;
    queue<pair<int, int>> q;
    q.push({s, 0});
    v[s][0] = 1;

    while (!q.empty())
    {
        auto tr = q.front();
        q.pop();
        if (tr.first == t && tr.second % 3 == 0)
            return cout << tr.second / 3 << endl, 0;
        for (int i : g[tr.first])
            if (!v[i][(tr.second + 1) % 3])
                q.push({i, tr.second + 1}), v[i][(tr.second + 1) % 3] = 1;
    }
    cout << -1 << endl;
}