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 <vector>
#include <queue>

using namespace std;

typedef pair<int, long> P;
typedef vector< vector<int> > V;

void bfs(queue<P>& q, V& to, V&ans) {
    while (!q.empty()) {
        P p = q.front();
        q.pop();
        int u = p.first;
        int c = p.second + 1;
        for (auto i : to[u]) {
            if (ans[c%3][i] != -1) continue;
            ans[c%3][i] = c;
            q.push(P(i, c));
        }
    }
}

int main() {
    int n, m, s, t;
    cin >> n >> m;
    V to(n+1);
    V ans(3, vector<int>(n+1, -1));
    for (int i=0; i<m; i++) {
        int u, v;
        cin >> u >> v;
        to[u].push_back(v);
    }
    cin >> s >> t;

    queue<P> q;
    q.push(P(s, 0));

    bfs(q, to, ans);

    if (ans[0][t] != -1) ans[0][t] /= 3;

    cout << ans[0][t] << endl;
}
