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 <bits/stdc++.h>
using namespace std;

int visit[100005][3];
vector<int> X[100005];

queue< pair<int, int> > Q;

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(NULL), cout.tie(NULL);

    int N, M;
    cin >> N >> M;

    while(M--)
    {
        int u, v;
        cin >> u >> v;

        X[u].push_back(v);
    }

    int s, t;
    cin >> s >> t;

    Q.push({s, 0});
    visit[s][0] = 1;

    while(!Q.empty())
    {
        int indx = Q.front().first, f = Q.front().second;
        Q.pop();

        for(auto x : X[indx])
        {
            if(visit[x][(f+1)%3])
                continue;
            
            visit[x][(f+1)%3] = visit[indx][f] + 1;
            Q.push({x, (f+1)%3});
        }
    }

    cout << (visit[t][0] == 0 ? -1 : (visit[t][0] - 1) / 3) << "\n";
    return 0;
}
