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 <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <tuple>
using namespace std;

int main() {
    int n, m; cin >> n >> m;
    using Graph = vector<vector<int>>;
    Graph g(n);
    while (m--) {
        int u, v; cin >> u >> v; u--, v--;
        g[u].emplace_back(v);
    }
    int s, t; cin >> s >> t; s--, t--;
    vector<vector<int>> dp(n, vector<int>(3, -1));
    queue<pair<int, int>> que;
    dp[s][0] = 0; que.emplace(s, 0);
    while (!que.empty()) {
        int u, r; tie(u, r) = que.front(); que.pop();
        int nr = (r + 1) % 3;
        for (int v: g[u]) if (dp[v][nr] == -1) {
            dp[v][nr] = dp[u][r] + 1;
            que.emplace(v, nr);
        }
    }
    if (~dp[t][0]) dp[t][0] /= 3;
    cout << dp[t][0] << endl;
    return 0;
}