Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

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

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

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

typedef long long int lli;
typedef pair<int, int> ii;
typedef vector<int> vi;
#define mkpr make_pair

const int N = 1e5 + 5;
const int M = 2e5 + 5;

int n, m;

vector<vector<ii>> edgeList(N);
vector<bool> vis(N, false);
vector<bool> inStack(N, false);
vector<bool> groupVis(M, false);

vector<vector<int>> vsGroups(N); 
vector<vector<int>> groupsVs(M);

map<int, int> colorReindexer; //colorReindexer[oldInd] gives newInd
int minUnusedIndex;

int main() {
	cin >> n >> m;

	minUnusedIndex = 0;
	int p, q, c;
	for(int i=0; i<m; i++) {
		cin >> p >> q >> c;
		p--; q--; c--;
		if(colorReindexer.find(c) == colorReindexer.end()) {
			colorReindexer.insert(mkpr(c, minUnusedIndex++));
		}
		edgeList[colorReindexer[c]].push_back(mkpr(p, q));
		edgeList[colorReindexer[c]].push_back(mkpr(q, p));
	}

	//Process by color
	int nGroup = 0;
	for(int i = 0; i < minUnusedIndex; i++) {
		vector<ii> companyProperty;
		
		for(ii e: edgeList[i]) {
			companyProperty.push_back(e);
			vis[e.first] = inStack[e.first] = false;
			vis[e.second] = inStack[e.second] = false;
		}

		sort(companyProperty.begin(), companyProperty.end());
		int unusedEdge = 0;
		stack<int> st;
		st.push(companyProperty[unusedEdge].first);
		st.push(companyProperty[unusedEdge].second);
		inStack[companyProperty[unusedEdge].first] = true;
		inStack[companyProperty[unusedEdge].second] = true;
		while(unusedEdge < companyProperty.size()) {
			while(!st.empty()) {
				int u = st.top(); st.pop();
				if(vis[u]) continue;
				vis[u] = true;
				
				vsGroups[u].push_back(nGroup);
				groupsVs[nGroup].push_back(u);

				auto starter = lower_bound(companyProperty.begin(), companyProperty.end(), mkpr(u, 0));
				auto ender = upper_bound(companyProperty.begin(), companyProperty.end(), mkpr(u, N));
				for(; starter != ender; starter++) {
					ii v = *starter;
					if(inStack[v.second]) continue;
					inStack[v.second] = true;
					st.push(v.second);
				}
			}
			nGroup++;

			while(unusedEdge < companyProperty.size() && vis[companyProperty[unusedEdge].first]) unusedEdge++;
		}
	}

	for(int i = 0; i < n; i++) vis[i] = groupVis[i] = false;

	int ans = 0;
	stack<int> st;
	stack<int> newSt;
	st.push(0);
	
	bool nReached = false;
	while(!st.empty() && !nReached) {
		stack<int> groupStack;
		while(!st.empty()) {
			int u = st.top(); st.pop();
			for(int g: vsGroups[u]) {
				if(!groupVis[g]) {
					groupStack.push(g);
					groupVis[g] = true;
				}
			}
		}

		while(!groupStack.empty()) {
			int g = groupStack.top(); groupStack.pop();
			for(int u: groupsVs[g]) {
				if(!vis[u]) {
					if(u == n-1) nReached = true;
					newSt.push(u);
					vis[u] = true;
				}
			}
		}

		ans++;
		while(!newSt.empty()) {
			st.push(newSt.top()); newSt.pop();
		}
	}

	cout << (nReached? ans:-1) << endl;

	return 0;
}