#include <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 7;

bool viz[N][3];
int d[N][3];
pair < int, int > q[3 * N];
vector < int > E[N];
int st, dr(-1);

void bfs() {
    while (st <= dr) {
        int x = q[st].first;
        int n = q[st++].second;
        int m = n + 1 - (n == 2 ? 3 : 0);
        for (int y : E[x]) {
            if (!viz[y][m]) {
                viz[y][m] = 1;
                d[y][m] = d[x][n] + 1;
                q[++dr] = {y, m};
            }
        }
    }
}

int main()
{
    int n, m, a, b, s, t;
    cin >> n >> m;
    while (m--)
        cin >> a >> b, E[a].push_back(b);
    cin >> s >> t;
    viz[s][0] = 1;
    q[++dr] = {s, 0};
    bfs();
    if (viz[t][0])
        cout << d[t][0] / 3;
    else
        cout << -1;
    return 0;
}

### 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 <bits/stdc++.h>

using namespace std;

const int N = 1e5 + 7;

bool viz[N][3];
int d[N][3];
pair < int, int > q[3 * N];
vector < int > E[N];
int st, dr(-1);

void bfs() {
    while (st <= dr) {
        int x = q[st].first;
        int n = q[st++].second;
        int m = n + 1 - (n == 2 ? 3 : 0);
        for (int y : E[x]) {
            if (!viz[y][m]) {
                viz[y][m] = 1;
                d[y][m] = d[x][n] + 1;
                q[++dr] = {y, m};
            }
        }
    }
}

int main()
{
    int n, m, a, b, s, t;
    cin >> n >> m;
    while (m--)
        cin >> a >> b, E[a].push_back(b);
    cin >> s >> t;
    viz[s][0] = 1;
    q[++dr] = {s, 0};
    bfs();
    if (viz[t][0])
        cout << d[t][0] / 3;
    else
        cout << -1;
    return 0;
}

```