#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define M (1000000007)
#define N (100010)
int n,m,f[N*10][3],dp[N][3];
vector<int>G[N];

int main(){
	scanf("%d%d",&n,&m);
	int x,y;
	for (int i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		G[x].push_back(y);
	}
	scanf("%d%d",&x,&y);
	f[1][0]=x; f[1][1]=0; f[1][2]=0;
	int l=0,r=1; memset(dp,0x7f7f,sizeof(dp));
	while (l<r){
		l++;
		if (f[l][0]==y&&f[l][1]==0){
			printf("%d",f[l][2]);
			return 0;
		}
		int now=f[l][0],cnt=f[l][2]+(f[l][1]+1)/3,t=(f[l][1]+1)%3;
		for (int i=0;i<G[now].size();i++){
			int nx=G[now][i];
			if (dp[nx][t]<=cnt) continue;
			dp[nx][t]=cnt;
			f[++r][0]=nx; f[r][1]=t; f[r][2]=cnt;
		}
	}
	printf("-1");
}

### Prompt

Please create a solution in CPP 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 ll long long
#define M (1000000007)
#define N (100010)
int n,m,f[N*10][3],dp[N][3];
vector<int>G[N];

int main(){
	scanf("%d%d",&n,&m);
	int x,y;
	for (int i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		G[x].push_back(y);
	}
	scanf("%d%d",&x,&y);
	f[1][0]=x; f[1][1]=0; f[1][2]=0;
	int l=0,r=1; memset(dp,0x7f7f,sizeof(dp));
	while (l<r){
		l++;
		if (f[l][0]==y&&f[l][1]==0){
			printf("%d",f[l][2]);
			return 0;
		}
		int now=f[l][0],cnt=f[l][2]+(f[l][1]+1)/3,t=(f[l][1]+1)%3;
		for (int i=0;i<G[now].size();i++){
			int nx=G[now][i];
			if (dp[nx][t]<=cnt) continue;
			dp[nx][t]=cnt;
			f[++r][0]=nx; f[r][1]=t; f[r][2]=cnt;
		}
	}
	printf("-1");
}

```