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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import Control.Monad
import Data.IntMap (IntMap, (!))
import qualified Data.IntMap as IM
import qualified Data.ByteString.Char8 as B
import Data.List
import Data.Bool
import Data.Maybe

solve :: Int -> IntMap [(Int,Int)] -> Bool
solve n nbr = isJust $ foldM f IM.empty [1..n] where
  f pos i
    | i `IM.member` pos = Just pos
    | otherwise = g i 0 pos
    where
      g i pi pos = foldM h (IM.insert i pi pos) (nbr!i) where
        h pos (j, d)
          | j `IM.member` pos =
              bool Nothing (Just pos) (pos!j == pj)
          | otherwise = g j pj pos
          where pj = pi + d

main :: IO ()
main = do
  [n,m] <- map read.words <$> getLine
  edge <- replicateM m readIntsLn
  let
    nbr = foldr f emps edge
    f [l,r,d] nbr
      = IM.adjust ((r,d):) l $ IM.adjust ((l,-d):) r nbr
    emps = IM.fromDistinctAscList $ zip [1..n] $ repeat [] 
  putStrLn $ bool "No" "Yes" $ solve n nbr

readIntsLn
  = unfoldr (B.readInt . B.dropWhile(==' ')) <$> B.getLine