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 <string>
#include <vector>
#include <list>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <functional>
#include <utility>
#include <fstream>

using namespace std;

#define ll long long
#define INF 2000000000
#define LINF 9000000000000000000
#define rep(i,n) for(int i=0;i<(int)n;i++)
#define dum(x) cout<<#x<<'='<<x<<endl
#define dum2(x,y) cout<<#x<<'='<<x<<' '<<#y<<'='<<y<<'\n'<<endl
#define dum3(x,y,z) cout<<#x<<'='<<x<<' '<<#y<<'='<<y<<' '<<#z<<'='<<z<<'\n'<<endl
#define darr(arr,n) rep(i,(n-arr)){cout<<*(arr+i)<<' ';}cout<<endl
#define ddarr(arr,a,b) rep(i,a){rep(j,b){cout<<*(arr+i*b+j)<<' ';}cout<<endl;}
#define ddd ifstream in("Text.txt");cin.rdbuf(in.rdbuf());

template<class Abel> struct unionfind {
    vector<int> par;
    vector<int> sizes;
    vector<Abel> wei;

    unionfind(int N, Abel uni = 0) : par(N), sizes(N, 1), wei(N, uni) {
        for (int i = 0; i < N; i++) par[i] = i;
    }

    int parfind(int x) {
        if (par[x] == x)return x;
        wei[x] += wei[par[x]];
        return par[x] = parfind(par[x]);
    }

    Abel weight(int x) {
        parfind(x);
        return wei[x];
    }

    void unite(int x, int y, Abel w) {
        w += weight(x); w -= weight(y);
        x = parfind(x); y = parfind(y);
        if (x == y) return;
        if (sizes[x] < sizes[y]) swap(x, y), w = -w;
        par[y] = x;
        wei[y] = w;
        sizes[x] += sizes[y];
    }

    bool same(int x, int y) {
        return parfind(x) == parfind(y);
    }
    Abel diff(int x, int y) {
        return weight(y) - weight(x);
    }
    int size(int x) {
        return sizes[parfind(x)];
    }
};

int main() {
    int N, M;
    cin >> N >> M;
    unionfind<int> uf(N);
    for (int i = 0; i < M; ++i) {
        int L, R, D;
        cin >> L >> R >> D;
        --L, --R;
        if (uf.same(L, R)) {
            int diff = uf.diff(L, R);
            if (diff != D) {
                cout << "No" << endl;
                return 0;
            }
        }
        else {
            uf.unite(L, R, D);
        }
    }
    cout << "Yes" << endl;
}