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<cstdio>
#include<iostream>
#include<vector>
#include<queue>
#include<bitset>
#include<string>
#include<stack>
#include<set>
#include<map>
#include<cstring>
#include<complex>
#include<cmath>
#include<algorithm>
#include<list>
#include<functional>
#define mp make_pair
#define X first
#define Y second
#define INF 987654321
#define PI 3.14159265358979323846264
#define MOD 1000000007LL
#define fup(i,a,b,c) for(int (i)=(a);(i)<=(b);(i)+=(c))
#define fdn(i,a,b,c) for(int (i)=(a);(i)>=(b);(i)-=(c))
#define MEM0(a) memset(a,0,sizeof(a));
#define MEM_1(a) memset(a,-1,sizeof(a));
#define ALL(a) a.begin(),a.end()
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int, int> Pi;
typedef pair<ll, ll> Pll;
typedef pair<double, double> Pd;
typedef vector<int> Vi;
typedef vector<ll> Vll;
typedef vector<double> Vd;
typedef vector<Pi> VPi;
typedef vector<Pll> VPll;
typedef vector<Pd> VPd;
typedef tuple<int, int, int> iii;
typedef vector<iii> Viii;
typedef complex<double> base;
ll POW(ll a, ll b) { if (b == 0)return 1; if (b == 1)return a; if (b & 1)return (a*POW(a, b - 1)) % MOD; ll t = POW(a, b / 2); return (t*t) % MOD; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
ll lcm(ll a, ll b) { if (a == 0 || b == 0)return gcd(a, b); return a*b / gcd(a, b); }
int dx[] = { 0,-1,1,0 }, dy[] = { 1,0,0,-1 };//동북남서
int ddx[] = { 0,0,1,1,1,-1,-1,-1 }, ddy[] = { 1,-1,1,0,-1,1,0,-1 };

int n, m;
int parent[100001];
int leng[100001];
int find(int a)
{
	if (parent[a] < 0)return a;
	find(parent[a]);
	leng[a] += leng[parent[a]];
	return parent[a] = find(parent[a]);
}
void merge(int a, int b,int d)
{
	leng[b] = leng[a] + d;
	a = find(a), b = find(b);
	if (a != b)
	{
		parent[a] += parent[b];
		parent[b] = a;
	}
}
int main() {
	MEM_1(parent);
	scanf("%d%d", &n, &m);
	fup(i, 0, m - 1, 1)
	{
		int l, r, d;
		scanf("%d%d%d", &l, &r, &d);
		if (find(l)!=find(r))
		{
			d += leng[l] - leng[r];
			merge(find(l), find(r), d);
		}
		else
		{
			if (leng[l] + d != leng[r])
			{
				puts("No");
				return 0;
			}
		}
	}
	puts("Yes");
}