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 Go (1.6) (judge verdict: Accepted)

package main
import (
  "bufio"
  "os"
  "strconv"
  "strings"
  "fmt"
  "container/heap"
)

type Item struct {
  Priority int
  Entry interface{}
}
type Heap []*Item
func (H Heap) Len() int { return len(H) }
func (H Heap) Less(i, j int) bool { return H[i].Priority < H[j].Priority }
func (H Heap) Swap(i, j int) { H[i], H[j] = H[j], H[i] }
func (H *Heap) Pop() interface{} {
  Old := *H
  n := len(Old)
  item := Old[n - 1]
  *H = Old[0:n - 1]
  return item
}
func (H *Heap) Push(x interface{}) {
  item := x.(*Item)
  *H = append(*H, item)
}
type PriorityQueue struct {
  Comparator func(e interface{}) int
  Items Heap
}
func (PQ *PriorityQueue) Push(e interface{}) {
  item := Item{PQ.Comparator(e), e}
  heap.Push(&PQ.Items, &item)
}
func (PQ *PriorityQueue) Pop() interface{} {
  return heap.Pop(&PQ.Items).(*Item).Entry
}
func (PQ PriorityQueue) Len() int { return len(PQ.Items) }
func InitPQ(F func(interface{}) int) PriorityQueue {
  var H Heap
  return PriorityQueue{F, H}
}

func Solve() {
  var N, M, p, q, c, d, a int
  var ok bool
  var pq []int
  NextInt(&N, &M)
  V := make([]map[int][]int, N)
  for i := range V { V[i] = make(map[int][]int) }
  G := make([][][]int, N + 2 * M)
  for i := range G { G[i] = make([][]int, 0) }
  for i := 0; i < M; i++ {
    NextInt(&p, &q, &c)
    p--; q--
    if _, ok = V[p][c]; !ok { V[p][c] = make([]int, 0) }
    V[p][c] = append(V[p][c], i)
    if _, ok = V[q][c]; !ok { V[q][c] = make([]int, 0) }
    V[q][c] = append(V[q][c], i)
    j := N + 2 * i
    G[p] = append(G[p], []int{j, 1})
    G[j] = append(G[j], []int{p, 0})
    G[j] = append(G[j], []int{j + 1, 0})
    G[j + 1] = append(G[j + 1], []int{j, 0})
    G[j + 1] = append(G[j + 1], []int{q, 0})
    G[q] = append(G[q], []int{j + 1, 1})
  }
  for _, v := range V {
    for k := range v {
      for i := range v[k][1:] {
        p, q = 2 * v[k][i], 2 * v[k][i + 1]
        G[N + p] = append(G[N + p], []int{N + q, 0})
        G[N + q] = append(G[N + q], []int{N + p, 0})
      }
    }
  }
  D := make([]int, N + 2 * M)
  for i := range D[1:] { D[i + 1] = N + M }
  F := func(e interface{}) int { return e.([]int)[0] }
  PQ := InitPQ(F)
  PQ.Push([]int{0, 0})
  for 0 < PQ.Len() {
    pq = PQ.Pop().([]int)
    d, p = pq[0], pq[1]
    if d > D[p] { continue }
    for _, adj := range G[p] {
      q, a = adj[0], adj[1]
      if d + a >= D[q] { continue }
      D[q] = d + a
      PQ.Push([]int{d + a, q})
    }
  }
  if D[N - 1] >= N { D[N - 1] = -1 }
  Write(D[N - 1])
}

func main() {
  Solve()
  Output()
}

var reader = bufio.NewReaderSize(os.Stdin, 1000000)
var writer = bufio.NewWriter(os.Stdout)
func NextLine() string {
  buffer := make([]byte, 0)
  for true {
    line, isPrefix, err := reader.ReadLine()
    if err != nil { panic(err) }
    buffer = append(buffer, line...)
    if !isPrefix { break }
  }
  return string(buffer)
}
func NextInt(A ...*int) {
  L := strings.Split(NextLine(), " ")
  for i, a := range A {
    *a, _ = strconv.Atoi(L[i])
  }
}
func NextIntVec(A *[]int) {
  L := strings.Split(NextLine(), " ")
  (*A) = make([]int, len(L))
  for i, l := range L {
    (*A)[i], _ = strconv.Atoi(l)
  }
}
func NextFloat(A ...*float64) {
  L := strings.Split(NextLine(), " ")
  for i, a := range A {
    *a, _ = strconv.ParseFloat(L[i], 64)
  }
}
func NextFloatVec(A *[]float64) {
  L := strings.Split(NextLine(), " ")
  (*A) = make([]float64, len(L))
  for i, l := range L {
    (*A)[i], _ = strconv.ParseFloat(l, 64)
  }
}
func Write(s interface{}) {
  fmt.Fprintln(writer, s)
}
func WriteIntVec(A []int) {
  S := make([]string, len(A))
  for i, a := range A {
    S[i] = strconv.Itoa(a)
  }
  Write(strings.Join(S, " "))
}
func Output() {
  _ = writer.Flush()
}
func MaxInt(A ...int) int {
  max := A[0]
  for _, a := range A {
    if max < a { max = a }
  }
  return max
}
func MinInt(A ...int) int {
  min := A[0]
  for _, a := range A {
    if a < min { min = a }
  }
  return min
}
func SumInt(A ...int) int {
  sum := 0
  for _, a := range A {
    sum += a
  }
  return sum
}
func AbsInt(x int) int {
  if x < 0 { x = -x }
  return x
}