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: Time Limit Exceeded)

import java.io.{BufferedReader, InputStream, InputStreamReader}
import java.util.StringTokenizer

import scala.collection.mutable
import scala.util.Sorting
import math.{abs, max, min}
import scala.collection.mutable.ArrayBuffer

object Main {
  val MOD = 1000000007
  def solve(): Unit = {
    val sc = new InputReader(System.in)
    val N = sc.nextInt()
    val M = sc.nextInt()

    type Node = Int
    type Edge = (Node, Int)
    val E = Array.ofDim[ArrayBuffer[Edge]](N)
    (0 until M) foreach { _ =>
      val l = sc.nextInt() - 1
      val r = sc.nextInt() - 1
      val d = sc.nextInt()
      if (E(l) == null) E(l) = ArrayBuffer.empty
      if (E(r) == null) E(r) = ArrayBuffer.empty
      E(l) += ((r, d))
      E(r) += ((l, -d))
    }

    val d = Array.ofDim[Int](N)
    type Elm = (Node, Int)
    val que = mutable.Queue.empty[Elm]

    (0 until N) foreach { i =>
      if (d(i) == 0) {
        visit(i, 0)
        while(que.nonEmpty) {
          val (cost, v) = que.dequeue()
          if (d(v) == 0) {
            d(v) = cost
            visit(v, cost)
          }
        }
      }
    }

    def visit(v: Int, cost0: Int): Unit = {
      if (E(v) != null) {
        E(v).indices foreach { i =>
          val (u, cost) = E(v)(i)
          if (d(u) == 0) que.enqueue((cost0 + cost, u))
        }
      }
    }

    var valid = true
    (0 until N) foreach { v =>
      if (E(v) != null) {
        E(v).indices foreach { j =>
          val (u, cost) = E(v)(j)
          valid &= d(v) + cost == d(u)
        }
      }
    }

    if (valid) println("Yes") else println("No")
  }


  def main(args: Array[String]): Unit = {
    solve()
  }


  class InputReader(val stream: InputStream) {
    private val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private var tokenizer: StringTokenizer = null

    def next: String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = next.toInt
    def nextLong(): Long = next.toLong
    def nextChar(): Char = next.charAt(0)
  }

}