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;
const int N = (int)1e5 + 50, INF = (int)1e9;

int n, m;
map<int, vector<int> > G[N];
set<int> rem[N];
int dis[N];

deque<P> que;

void dfs(int v, int color, int d) {
//    printf("in %d , color: %d\n", v, color);
    rem[v].erase(color);
    if(d < dis[v]) {
        dis[v] = d;
        que.push_back({v, color});
    }
    for(int nxt : G[v][color]) {
        if(rem[nxt].find(color) != rem[nxt].end()) {
//            printf("from %d to %d\n", v, nxt);
            dfs(nxt, color, d);
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> m;
    fill(dis, dis + n, INF);
    for(int i = 0; i < m; i++) {
        int a, b, c; cin >> a >> b >> c; a--, b--, c--;
        G[a][c].push_back(b);
        G[b][c].push_back(a);
        rem[a].insert(c);
        rem[b].insert(c);
    }
    while(!rem[0].empty()) dfs(0, *rem[0].begin(), 1);
//    for(int i = 0; i < n; i++) {
//        printf("dis %d: %d\n", i + 1, dis[i]);
//    }
    while(!que.empty()) {
        P p = que.front();
        int v = p.first, c = p.second;
        que.pop_front();
        while(!rem[v].empty()) {
            dfs(v, *rem[v].begin(), dis[v] + 1);
        }
    }
    if(dis[n-1] == INF) dis[n-1] = -1;
    cout << dis[n - 1] << endl;
}