#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;

int main() {
    ios::sync_with_stdio(false);
    int n, m, u, v, s, t, d[3][100005] = {};
    vector<int> g[100005] = {};
    cin >> n >> m;
    while (m--) cin >> u >> v, g[u].push_back(v);
    cin >> s >> t;
    queue<pii> q;
    q.push({0, s});
    d[0][s] = 1;
    while (!q.empty()) {
        int x = q.front().first, y = q.front().second;
        q.pop();
        for (int i : g[y]) {
            if (!d[(x + 1) % 3][i]) d[(x + 1) % 3][i] = d[x][y] + 1, q.push({(x + 1) % 3, i});
        }
    }
    if (d[0][t]) cout << (d[0][t] - 1) / 3;
    else cout << -1;
}

### Prompt

Create a solution in CPP for the following problem:

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

### Response

```cpp
#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <cmath>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int mod = 1000000007;

int main() {
    ios::sync_with_stdio(false);
    int n, m, u, v, s, t, d[3][100005] = {};
    vector<int> g[100005] = {};
    cin >> n >> m;
    while (m--) cin >> u >> v, g[u].push_back(v);
    cin >> s >> t;
    queue<pii> q;
    q.push({0, s});
    d[0][s] = 1;
    while (!q.empty()) {
        int x = q.front().first, y = q.front().second;
        q.pop();
        for (int i : g[y]) {
            if (!d[(x + 1) % 3][i]) d[(x + 1) % 3][i] = d[x][y] + 1, q.push({(x + 1) % 3, i});
        }
    }
    if (d[0][t]) cout << (d[0][t] - 1) / 3;
    else cout << -1;
}

```