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>
using namespace std;

typedef pair<int,int> P;
typedef pair<int, P> pp;
struct edge{P to; int cost;};

int N, M;
map<P, vector<edge>> G;
map<P, int> d;

void dijkstra(int s) {
    priority_queue<pp, vector<pp>, greater<pp> > que;
    d[P(s, 0)] = 0;
    que.push(pp(0, P(s, 0)));

    while(!que.empty()) {
        pp p = que.top(); que.pop();
        P v = p.second;
        if(d[v] < p.first) continue;
        for(edge e : G[v]) {
            if(d.find(e.to) == end(d) || d[e.to] > d[v] + e.cost) {
                d[e.to] = d[v] + e.cost;
                que.push(pp(d[e.to], e.to));
            }
        }
    }
}

int main() {
    cin >> N >> M;
    for(int i=0; i<M; ++i) {
        int p, q, c;
        cin >> p >> q >> c;
        --p; --q;
        if(G.find(P(p, c)) == end(G)) {
            G[P(p, 0)].push_back({P(p, c), 1});
            G[P(p, c)].push_back({P(p, 0), 1});
        }
        if(G.find(P(q, c)) == end(G)) {
            G[P(q, 0)].push_back({P(q, c), 1});
            G[P(q, c)].push_back({P(q, 0), 1});
        }
        G[P(p, c)].push_back({P(q, c), 0});
        G[P(q, c)].push_back({P(p, c), 0});
    }

    dijkstra(0);
    if(d.find(P(N - 1, 0)) == end(d)) cout << -1 << endl;
    else cout << d[P(N - 1, 0)] / 2 << endl;
}