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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

#ifdef DEBUG
#define NDEBUG
#include "cout11.h"
#endif
#undef NDEBUG
#include <cassert>

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> llll;
typedef pair<double,double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;

#define sz(a)  int((a).size())
#define pb  push_back
#define FOR(var,from,to) for(int var=(from);var<=(to);++var)
#define rep(var,n)  for(int var=0;var<(n);++var)
#define rep1(var,n)  for(int var=1;var<=(n);++var)
#define repC2(vari,varj,n)  for(int vari=0;vari<(n)-1;++vari)for(int varj=vari+1;varj<(n);++varj)
#define ALL(c)  (c).begin(),(c).end()
#define RALL(c)  (c).rbegin(),(c).rend()
#define tr(i,c)  for(auto i=(c).begin(); i!=(c).end(); ++i)
#define found(s,e)  ((s).find(e)!=(s).end())
#define mset(arr,val)  memset(arr,val,sizeof(arr))
#define mid(x,y) ((x)+((y)-(x))/2)
#define IN(x,a,b) ((a)<=(x)&&(x)<=(b))

class UnionFind {
  vector<int> data;
 public:
  UnionFind(int size) : data(size, -1) { }
  bool unionSet(int x, int y) {
    x = root(x); y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y]; data[y] = x;
    }
    return x != y;
  }
  bool findSet(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  int size(int x) { return -data[root(x)]; }
};


int N, M;
vvi f;
map<int, vi> mv;
vector<vector<ii>> arc;

bool sub(int rt){
    int p1 = mv[rt][0];
    set<int> vis;
    vis.insert(p1);

    vi loc(N);
    vector<bool> locset(N, false);

    loc[p1] = 0; // 基準
    locset[p1] = true;

    queue<int> q;
    for (auto p: arc[p1]) {
        loc[p.first] = loc[p1] + p.second;
        locset[p.first] = true;
        q.push(p.first);
    }

    while (!q.empty()) {
        int x = q.front(); q.pop();
        if (found(vis, x)) continue;

        for (auto p : arc[x]) {
            if (locset[p.first]) {
                if (loc[p.first] != loc[x] + p.second) return false;
            } else {
                loc[p.first] = loc[x] + p.second;
                locset[p.first] = true;
                if (!found(vis, p.first)) q.push(p.first);
            }
        }
    }
#ifdef DEBUG
    for (int i : mv[rt]) {
        printf("(%d:%d)%c", i, loc[i], locset[i]?' ':'x');
    }
    printf("\n");
#endif

    return true;
}

bool solve(){
    arc.resize(N);

    UnionFind uf(N);
    rep(i,M) {
        int li = f[i][0], ri = f[i][1], di = f[i][2];
        uf.unionSet(li, ri);
        arc[li].pb(ii(ri, di));   // ri = li + di
        arc[ri].pb(ii(li, -di));  // li = ri - di
    }

    rep(i,N) {
        int rt = uf.root(i);
        mv[rt].pb(i);
    }

    for (auto p: mv){
        int rt = p.first;
        if (!sub(rt)) return false;
    }

    return true;
}

int main() {
    cin>>N>>M;
    f.resize(M);
    rep(i,M){
        int li, ri, di; cin >> li >> ri >> di;
        f[i] = { li-1, ri-1, di };
    }
    cout << (solve() ? "Yes" : "No") << endl;
    return 0;
}