Problem: Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.

The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.

You can change trains at a station where multiple lines are available.

The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.

Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.

Constraints

* 2 \leq N \leq 10^5
* 0 \leq M \leq 2×10^5
* 1 \leq p_i \leq N (1 \leq i \leq M)
* 1 \leq q_i \leq N (1 \leq i \leq M)
* 1 \leq c_i \leq 10^6 (1 \leq i \leq M)
* p_i \neq q_i (1 \leq i \leq M)

Input

The input is given from Standard Input in the following format:


N M
p_1 q_1 c_1
:
p_M q_M c_M


Output

Print the minimum required fare. If it is impossible to get to station N by subway, print `-1` instead.

Examples

Input

3 3
1 2 1
2 3 1
3 1 2


Output

1


Input

8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5


Output

2


Input

2 0


Output

-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define maxn 500086

using namespace std;

int n, m;
map<pair<int, int>, int> mp;
int x, y, z;
int cnt;

vector<pair<int, int> > v[maxn]; 

inline void addEdge(int x, int y, int z){
	v[x].push_back({y, z}), v[y].push_back({x, z});
}

priority_queue<pair<int, int>, vector<pair<int, int> >, greater<pair<int, int> > > q;
bool vis[maxn];
int dis[maxn];

int main(){
	scanf("%d%d", &n, &m), cnt = n;
	while(m--){
		scanf("%d%d%d", &x, &y, &z);
		if(!mp[{x, z}]) mp[{x, z}] = ++cnt;
		if(!mp[{y, z}]) mp[{y, z}] = ++cnt;
		int xx = mp[{x, z}], yy = mp[{y, z}];
		addEdge(x, xx, 1), addEdge(xx, yy, 0), addEdge(yy, y, 1);
	}
	memset(dis, 0x3f, sizeof(dis));
	dis[1] = 0, q.push({dis[1], 1});
	for(int i = 1;i <= cnt;i++){
		while(!q.empty() && vis[q.top().second]) q.pop();
		if(q.empty()) break;
		int x = q.top().second;q.pop();
		vis[x] = true;
		for(int j = 0;j < v[x].size();j++){
			int to = v[x][j].first, co = v[x][j].second;
			if(dis[to] > dis[x] + co){
				dis[to] = dis[x] + co;
				q.push({dis[to], to});
			}
		}
	}
	printf("%d", dis[n] == 0x3f3f3f3f ? -1 : (dis[n] / 2));
}