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 <cstdio>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <algorithm>

using namespace std;

int d[100000];
set <int> s[100000];
map <int, vector<int> > mp[100000];

void dfs(int x, int y, int z, queue <int> &q) {
    int i;
    
    s[x].insert(y);
    
    if (d[x] > z) {
        d[x] = z;
        q.push(x);
    }
    
    for (i = 0; i < mp[x][y].size(); i++) {
        int nx = mp[x][y][i];
        
        if (!s[nx].count(y)) dfs(nx, y, z, q);
    }
}

int main() {
    int n, m, i;
    queue <int> q;
    
    scanf("%d %d", &n, &m);
    
    for (i = 0; i < m; i++) {
        int x, y, z;
        
        scanf("%d %d %d", &x, &y, &z);
        
        x--;
        y--;
        
        mp[x][z].push_back(y);
        mp[y][z].push_back(x);
    }
    
    for (i = 0; i < n; i++) d[i] = 1e9;
    
    d[0] = 0;
    q.push(0);
    
    while (!q.empty()) {
        int x = q.front();
        map <int, vector <int> >::iterator it;
        
        q.pop();
        
        if (x == n - 1) {
            printf("%d\n", d[n - 1]);
            
            return 0;
        }
        
        for (it = mp[x].begin(); it != mp[x].end(); it++) {
            if (!s[x].count(it->first)) dfs(x, it->first, d[x] + 1, q);
        }
    }
    
    puts("-1");
    
    return 0;
}