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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <algorithm>
#include <limits.h>
#include <vector>
#include <map>
#include <stack>
#include <queue>
#include <math.h>
#define ll long long
#define MSET(x) memset((x), 0, sizeof((x)))
#define MMSET(x) memset((x), -1, sizeof((x)))
#define rep(i, a, b) for ( int (i) = a; (i) < (b) ; (i) ++ )
#define repA(i, a, b) for ( int (i) = a ;  (i) <= (b) ; (i) ++ )
#define repD(i, a, b) for ( int (i) = a; (i) >= (b) ; (i) -- )
#define fi first
#define se second
#define pb push_back
#define mfind(m, item) ((m).find((item)) != (m).end())
#define mp(a,b) make_pair((a),(b))
#define mt(a,b,c) make_pair(make_pair((a),(b)), (c))
#define tfi first.first
#define tse first.second
#define tth second
#define print cout << endl
#define tri pair<pair<int, int>, int>
using namespace std;

int n,m;
int fa[100005], cnt[100005];
void find(int x){
    if ( fa[x] == x ) return;
    find(fa[x]);
    cnt[x] += cnt[fa[x]]; fa[x] = fa[fa[x]];
}

int main(){
    while ( cin >> n >> m ) {
        bool has = false;
        MSET(fa); MSET(cnt);
        for ( int i = 0 ; i <= n ; ++ i ) fa[i] = i;
        for ( int i = 0, x, y, z ; i < m ; ++ i ) {
            scanf("%d %d %d", &x, &y, &z);
            find(x); find(y);
            if ( fa[x] == fa[y] ) {
                if ( cnt[y] != cnt[x] + z ) has = true;
            }
            else {
                cnt[fa[y]] = z;
                fa[fa[y]] = x;
            }
        }
        cout << (has ? "No" : "Yes" ) << endl;
    }
}