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: Accepted)

#include <bits/stdc++.h>
#define Maxn 400007
using namespace std;
int n,m;
int cnt=0,last[Maxn],pre[Maxn],other[Maxn],val[Maxn];
long long mn,mx;
long long t[Maxn];
bool vis[Maxn];
bool ans=true;
void insert(int u,int v,int t)
{
	other[++cnt]=v,pre[cnt]=last[u],last[u]=cnt;
	val[cnt]=t;
}
void dfs(int u,int fa)
{
	for (int q=last[u];q;q=pre[q])
	{
		int v=other[q];
		if (v!=fa)
		{
			if (!vis[v])
			{
				vis[v]=true;
				t[v]=t[u]+val[q];
				mx=max(mx,t[v]);
				mn=min(mn,t[v]);
				dfs(v,u);
			} else if (t[v]!=t[u]+val[q]) ans=false;
		}
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		int a,b,c;
		scanf("%d%d%d",&a,&b,&c);
		insert(a,b,c);
		insert(b,a,-c);
	}
	memset(vis,false,sizeof(vis));
	for (int i=1;i<=n;i++)
		if (!vis[i])
		{
			mn=0,mx=0;
			t[i]=0;
			dfs(i,-1);
			if (mx-mn>1000000000LL) ans=false;
		}
	if (ans) printf("Yes\n"); else printf("No\n");
	return 0;
}