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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (int)(n); i++)
#define REP(i,n) for (int i = 1; i < (int)(n); i++)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define vout(x) rep(i,x.size()) cout << x[i] << " "
template<class T> bool chmin(T &a, T b) {if(a>b) {a=b;return 1;}return 0;}
template<class T> bool chmax(T &a, T b) {if(a<b) {a=b;return 1;}return 0;}
using namespace std;
using vint = vector<int>;
using vvint = vector<vector<int>>;
using ll = long long;
using vll = vector<ll>;
using vvll = vector<vector<ll>>;
using P = pair<int,int>;
const int inf = 1e9;
const ll inf_l = 1e18;
const int MAX = 1e5;

// union-find
struct union_find {
    vector<int> data;

    union_find(int size) {data.resize(size,-1);}

    int find(int x) {
        if (data[x] < 0) {return x;}
        return data[x] = find(data[x]);
    }

    bool unite(int x, int y) {
        x = find(x); y = find(y);
        if (x == y) {return false;}
        if (data[x] > data[y]) swap(x,y);
        data[x] += data[y];
        data[y] = x;
        return true;
    }

    int size(int x) {return -data[find(x)];}
};

// dijkstra
int n;
vint d(MAX);

void dijkstra(int s, const vvint &cost) {
    priority_queue<P, vector<P>, greater<P>> q;
    rep(i,n) d[i] = inf;
    d[s] = 0;
    q.push(P(0,s));
    while (!q.empty()) {
        P e = q.top(); q.pop();
        int v = e.second;
        if (d[v] < e.first) continue;
        rep(i,n) {
            if (cost[v][i] == inf) continue;
            if (chmin(d[i],d[v]+cost[v][i])) {
                q.push(P(d[i],i));
            }
        }
    }
}

int main() {
    int m; cin >> n >> m;
    union_find uf(n);
    vvint cost(n,vint(n,inf));
    rep(i,m) {
        int l, r, dis; cin >> l >> r >> dis;
        l--; r--;
        if (uf.find(l) != uf.find(r)) {
            uf.unite(l,r);
            cost[l][r] = dis;
            cost[r][l] = dis;
        }
        else {
            dijkstra(l,cost);
            if (d[r] != dis) {cout << "No" << endl; return 0;}
        }
    }
    cout << "Yes" << endl;
}