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 long long ll;

struct st {
    int v, c, d;
};

int main() {
    constexpr int INF = 1e9 + 6;
    int n, m; cin >> n >> m;
    map<pair<int,int>,map<pair<int,int>, int>> G;
    map<pair<int, int>, int> ans;
    for (int i = 0; i < m; i++) {
        int p, q, c; cin >> p >> q >> c;
        p--; q--;
        G[{p, -1}][{p, c}] = 1;
        G[{q, -1}][{q, c}] = 1;
        G[{p, c}][{p, -1}] = 0;
        G[{q, c}][{q, -1}] = 0;
        G[{p, c}][{q, c}] = 0;
        G[{q, c}][{p, c}] = 0;
        ans[{p, -1}] = ans[{q, -1}] = ans[{p, c}] = ans[{q, c}] = INF;
    }
    deque<st> deq;
    deq.push_back({0,-1,0});
    ans[{0, -1}] = 0;
    while (!deq.empty()) {
        auto p = deq.front(); deq.pop_front();
        for (auto &q : G[{p.v,p.c}]) {
            if (ans[q.first] <= q.second + p.d) continue;
            if (q.second == 0) {
                deq.push_front({q.first.first,q.first.second,p.d});
            }
            else {
                deq.push_back({q.first.first,q.first.second,p.d + q.second});
            }
            ans[q.first] = q.second + p.d;
        }
    }
    if (ans.find({n-1,-1}) == ans.end() || ans[{n - 1, -1}] == INF) {
        puts("-1");
    }
    else {
        cout << ans[{n - 1, -1}] << endl;
    }
    return 0;
}