Problem:
Score : 400 points

Problem Statement

There are N people standing on the x-axis.
Let the coordinate of Person i be x_i.
For every i, x_i is an integer between 0 and 10^9 (inclusive).
It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people.
The i-th piece of information has the form (L_i, R_i, D_i).
This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds.

It turns out that some of these M pieces of information may be incorrect.
Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information.

Constraints

1 \leq N \leq 100 000

0 \leq M \leq 200 000

1 \leq L_i, R_i \leq N (1 \leq i \leq M)

0 \leq D_i \leq 10 000 (1 \leq i \leq M)

L_i \neq R_i (1 \leq i \leq M)

If i \neq j, then (L_i, R_i) \neq (L_j, R_j) and (L_i, R_i) \neq (R_j, L_j).

D_i are integers.

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 D_1
L_2 R_2 D_2
:
L_M R_M D_M

Output

If there exists a set of values (x_1, x_2, ..., x_N) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in Scala (2.11.7) (judge verdict: Runtime Error)

import scala.io.{StdIn => in}
import scala.collection.{mutable => mu}

object Main extends App {
  val Array(n, m) = in.readLine.split(" ").map(_.toInt)

  val INF = 1 << 30
  val value = Array.fill(n+1)(INF)
  case class Edge(node:Int, cost:Int)
  val graph = Array.fill(n+1)(mu.ListBuffer.empty[Edge])

  Iterator.continually(in.readLine)
    .take(m)
    .map(_.split(" ").map(_.toInt))
    .foreach(e => {
      val l = e(0) - 1
      val r = e(1) - 1
      val d = e(2)
      graph(l) += new Edge(r,d)
      graph(r) += new Edge(l,-d)
    })

  def bfs(node:Int):Unit = {
    if(graph(node).nonEmpty) {
      graph(node).foreach( edge => {
        if(value(edge.node) == INF) {
          value(edge.node) = value(node) + edge.cost
          bfs(edge.node)
        }
      })
    }
  }

  (0 until n).foreach( i => {
    if(value(i) == INF) {
      value(i) = 0
      bfs(i)
    }
  })

  val valid =
    (0 until n).forall( i => {
      graph(i).forall( edge => {
        value(i) + edge.cost == value(edge.node)
      })
    })

  val ans = if(valid) "Yes" else "No"
  println(ans)
}