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)

{-# OPTIONS_GHC -O #-}

import Control.Monad
import Control.Monad.ST
import Data.List
import qualified Data.Vector.Unboxed.Mutable as VM
import qualified Data.ByteString.Char8 as B
import qualified Data.IntMap as M

main = do
    (n:m:_)<-map read.words<$>getLine::IO [Int]
    lrds <- replicateM m $ unfoldr (B.readInt.B.dropWhile(==' ')) <$> B.getLine :: IO[[Int]]
    let grp = foldl (\m (l:r:d:_)->M.alter (ins l (-d)) r $ M.alter (ins r d) l m) M.empty lrds
    putStrLn $ if solve n grp lrds then "Yes" else "No"
--
ins ::Int -> Int -> Maybe [(Int,Int)] -> Maybe [(Int,Int)]
ins n w Nothing = Just [(n,w)]
ins n w (Just nws) = Just ((n,w):nws)
--
solve :: Int -> M.IntMap [(Int,Int)] -> [[Int]] -> Bool
solve n grp lrds = runST $ do
    ckd <- VM.replicate (n+1) False :: ST s (VM.STVector s Bool)
    dst <- VM.replicate (n+1) 0 :: ST s (VM.STVector s Int)
    fmap and $ forM [1..n] $ \i -> do
        d <- VM.unsafeRead dst i
        dfs grp ckd dst True (i,d)
--
dfs :: M.IntMap [(Int,Int)] -> VM.STVector s Bool -> VM.STVector s Int -> Bool -> (Int,Int) -> ST s Bool
dfs grp ckd dst ac (x,d) = do
    c <- VM.unsafeRead ckd x
    dd <- VM.unsafeRead dst x
    if c then return $ ac && dd==d
    else do
        VM.unsafeWrite ckd x True
        VM.unsafeWrite dst x d
        foldM (dfs grp ckd dst) ac $ map (\(u,v)->(u, v+d)) $ M.findWithDefault [] x grp