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;

const int K = 3;

int main() {
    int n, m;
    cin >> n >> m;
    vector<int> e[K*n];
    for (int i = 0; i < m; i++) {
        int u, v;
        cin >> u >> v;
        u--, v--;
        for (int j = 0; j < K; j++) {
            e[j*n+u].push_back((j+1)%K*n+v);
        }
    }
    int s, t;
    cin >> s >> t;
    s--, t--;

    int d[3*n];
    for (int i = 0; i < 3 * n; i++) {
        d[i] = -3;
    }
    d[s] = 0;

    queue<int> q;
    q.push(s);

    while (!q.empty()) {
        int x = q.front();
        q.pop();

        for (int y : e[x]) {
            if (d[y] == -3) {
                d[y] = d[x] + 1;
                q.push(y);
            }
        }
    }

    cout << d[t] / 3 << endl;

    return 0;
}
