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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iomanip>
using namespace std;
struct info
{
    int l,r,d;
}s[400005];
int n,m;
int flag = 1;
int dis[100005],visited[100005];
int first[400005],nexti[400005];
void dfs(int t)
{
    if(!flag)return;
    int k = first[t];
    while(k != -1)
    {
        if(!flag)return;
        if(!visited[s[k].r])
        {
            dis[s[k].r] = dis[s[k].l] + s[k].d;
            visited[s[k].r] = 1;
            dfs(s[k].r);
        }
        else if(s[k].d != dis[s[k].r] - dis[s[k].l])
        {
            flag = 0;
            return;
        }
        k = nexti[k];
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    memset(first,-1,sizeof(first));
    for(int i = 0;i < m;i ++)
    {
        scanf("%d%d%d",&s[i].l,&s[i].r,&s[i].d);
        nexti[i] = first[s[i].l];
        first[s[i].l] = i;
        s[i + m].r = s[i].l;
        s[i + m].l = s[i].r;
        s[i + m].d = -s[i].d;
        nexti[i + m] = first[s[i + m].l];
        first[s[i + m].l] = i + m;
    }
    for(int i = 1;i <= n;i ++)
    {
        if(!visited[i])
        {
            visited[i] = 1;
            dfs(i);
        }
        else break;
    }
    if(flag)printf("Yes");
    else printf("No");
}