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

#define ran 101101

int n, m;
vector<int> e[ran];
int S, T;
int d[ran][3];
int q[ran*2*3], l, r;

void upd(int pos, int rem, int val) {
	if(d[pos][rem]>=0) return;
	d[pos][rem] = val;
	q[r++] = pos;
	q[r++] = rem;
}

int main() {
	scanf("%d%d", &n, &m);
	for(int i=0;i<m;i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		e[x].push_back(y);
	}
	scanf("%d%d", &S, &T);
	
	memset(d, -1, sizeof d);
	l = r = 0;
	upd(S, 0, 0);
	while(l<r) {
		int cur = q[l++];
		int rem = q[l++];
		for(int i=0;i<(int)e[cur].size();i++) {
			int nxt = e[cur][i];
			upd(nxt, (rem+1)%3, d[cur][rem]+1);
		}
	}
	if(d[T][0]==-1) puts("-1");
	else printf("%d\n", d[T][0]/3);
	
	return 0;
}
### Prompt

Generate a CPP solution to 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;

#define ran 101101

int n, m;
vector<int> e[ran];
int S, T;
int d[ran][3];
int q[ran*2*3], l, r;

void upd(int pos, int rem, int val) {
	if(d[pos][rem]>=0) return;
	d[pos][rem] = val;
	q[r++] = pos;
	q[r++] = rem;
}

int main() {
	scanf("%d%d", &n, &m);
	for(int i=0;i<m;i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		e[x].push_back(y);
	}
	scanf("%d%d", &S, &T);
	
	memset(d, -1, sizeof d);
	l = r = 0;
	upd(S, 0, 0);
	while(l<r) {
		int cur = q[l++];
		int rem = q[l++];
		for(int i=0;i<(int)e[cur].size();i++) {
			int nxt = e[cur][i];
			upd(nxt, (rem+1)%3, d[cur][rem]+1);
		}
	}
	if(d[T][0]==-1) puts("-1");
	else printf("%d\n", d[T][0]/3);
	
	return 0;
}
```