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 <vector>
#include <string.h>
#include <stack>
#include <queue>
#include <algorithm>
#include <climits>
#include <cmath>
#include <map>
#include <set>
#include <assert.h>
#define REP(i,n) for(ll i=0;i<(n);i++)
#define MOD 1000000007
#define int long long
#ifdef int
const long long INF = LLONG_MAX / 10;
#else
const int INF = 1010101010;
#endif
using namespace std;
typedef long long ll;
typedef vector<vector<ll> > mat;
typedef pair<int, int> P;
//typedef pair<double, double> P;
const int dx[9] = {-1, 0, 0, 1, -1, -1, 1, 1, 0};
const int dy[9] = {0, -1, 1, 0, -1, 1, -1, 1, 0};

struct edge { int to, cost; };
vector<edge> G[101010];
int N, M;
int x[101010];
set<int> se;

bool dfs(int i)
{
  if (x[i] != LLONG_MIN) return true;
  for (edge e : G[i]) {
    if (x[e.to] != LLONG_MIN && x[e.to] != x[i] + e.cost) {
      return false;
    }
    x[e.to] = x[i] + e.cost;
    if (!dfs(e.to)) return false;
  }
  return true;
}

signed main()
{
  cin >> N >> M;
  REP(i,M) {
    int L, R, D;
    cin >> L >> R >> D;
    G[L].push_back((edge){R, D});
    G[R].push_back((edge){L, -D});
    se.insert(L);
    se.insert(R);
  }

  fill(x, x + 101010, LLONG_MIN);

  for (auto it=se.begin(); it!=se.end(); it++) {
    if (x[*it] == LLONG_MIN) {
      x[*it] = 0;
      if (!dfs(*it)) {
        cout << "No" << endl;
        return 0;
      }
    }
  }
  cout << "Yes" << endl;

  // REP(i,10) cout << x[i] << " ";
}