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 "bits/stdc++.h"
#define int long long
#define range(i, a, b) for(int i = (a); i < (b); i++)
#define rep(i, a) range(i, 0, a)
#define all(a) (a).begin(),(a).end()
using namespace std;
const int INF = sizeof(int) == sizeof(long long) ? 1000000000000000000LL : 1000000000;
const int MOD = 1000000007;
const long double EPS = 1e-8;
using vi = vector <int>;
using vvi = vector <vi>;

int n, m, sz;
vector <bool> used;
struct edge {int to; int dist;};
vector <vector <edge>> G;
map <int, int> pear, berry;
bool reach, mujun;

void dfs(int now, int goal, int expect, int sum) {
	if (mujun) return;
	if (now == goal) {
		reach = true;
		mujun = mujun || expect != sum;
	}

	rep (i, G[now].size()) {
		if (used[G[now][i].to]) continue;
		used[G[now][i].to] = true;
		dfs(G[now][i].to, goal, expect, sum + G[now][i].dist);
	}

	return;
}

signed main () {
	cin >> n >> m;
	G = vector <vector <edge>> (n);
	used = vector <bool> (n, false);
	int l[m], r[m], d[m];
	map <int, int> exist;

	bool flag = false;
	sz = 0;
	rep (i, m) {
		cin >> l[i] >> r[i] >> d[i];
		if (flag) continue;
		if (exist[l[i]] != 0 && exist[r[i]] != 0) {
			used = vector <bool> (sz, false);
			used[pear[l[i]]] = true;
			reach = false;
			mujun = false;
			dfs(pear[l[i]], pear[r[i]], d[i], 0);
			dfs(pear[r[i]], pear[l[i]], -d[i], 0);
			if (!reach || mujun) flag = true;
		} else {
			if (exist[l[i]] == 0) {
				exist[l[i]] = true;
				pear[l[i]] = sz;
				berry[sz] = l[i];
				sz++;
			} else {
				rep (j, G[pear[l[i]]].size()) if (G[pear[l[i]]][j].dist == d[i]) flag = true;
			}
			if (exist[r[i]] == 0) {
				exist[r[i]] = true;
				pear[r[i]] = sz;
				berry[sz] = r[i];
				sz++;
			} else {
				rep (j, G[pear[r[i]]].size()) if (G[pear[r[i]]][j].dist == -d[i]) flag = true;
			}
			G[pear[l[i]]].push_back({pear[r[i]], d[i]});
			G[pear[r[i]]].push_back({pear[l[i]], -d[i]});
		}
	//	cout << (flag ? "No" : "Yes") << endl;
	}

	cout << (flag ? "No" : "Yes") << endl;

	return 0;
}