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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#pragma region header
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rev(i, n) for(int i = (int)(n - 1); i >= 0; i--)
#define rev1(i, n) for(int i = (int)(n); i > 0; i--)
#define pb push_back
#define all(v) (v).begin(), (v).end()
#define resort(v) sort((v).rbegin(), (v).rend())
#define vi vector<int>
#define vvi vector<vector<int>>
#define vc vector<char>
#define vvc vector<vector<char>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
using ll = long long;
using P = pair<int, int>;
/* ----------------よく使う数字や配列----------------- */
int dx[] = {1,0,-1,0};
int dy[] = {0,1,0,-1};
constexpr ll mod = 1e9+7;
constexpr int inf = INT32_MAX/2;
constexpr ll INF = LLONG_MAX/2;
constexpr long double eps = DBL_EPSILON;
constexpr long double pi = 3.141592653589793238462643383279;
/* ----------------------end----------------------- */

/* --------------------テンプレート------------------ */
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
/* ----------------------end----------------------- */

/* --------------------ライブラリ-------------------- */
ll fact(int i) {       //階乗
    if (i == 0) return 1;
    return (fact(i - 1)) * i % mod;
}
ll gcd(ll a, ll b) {        //最大公約数
    if(b == 0) return a;
    return gcd(b, a % b); 
}
ll lcm(ll a, ll b) {      //最小公倍数
    return a / gcd(a, b) * b ;
}
int keta(ll n) {        //桁数を求める
    if(n == 0) return 1;
    int count = 0;
    while(n != 0) {
        n /= 10;
        count++;
    }
    return count;
}
ll ketasum(ll n) {    //各桁の和
    ll sum = 0;
    while(n != 0) {
        sum += n % 10;
        n /= 10;
    }
    return sum;
}
bool isprime(int n) { //素数判定
    if(n==1)return false;
    if(n==2)return true;
    for(int i = 2;i<=sqrt(n);i++) {
        if(n%i==0) return false;
    }
    return true;
}
/* ----------------------end----------------------- */
#pragma endregion
struct edge{int to,cost;};
vector<vector<edge>> G(100100);
struct UnionFind {
    vector<int> par, size;
    UnionFind(int x) {
        par.resize(x);
        size.resize(x, 1);
        for(int i = 0; i < x; i++) {
            par[i] = i;
        }
    }
    int find(int x) {
        if (par[x] == x)
            return x;
        return par[x] = find(par[x]);
    }
    bool same(int x, int y) {
        return find(x) == find(y);
    }
    int consize(int x) {
        return size[find(x)];
    }
    bool unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y)
            return false;
        if (size[x] < size[y]) {
            par[x] = y;
            size[y] += size[x];
        }
        else {
            par[y] = x;
            size[x] += size[y];
        }
        return true;
    }
};
map<P,int> mp;
bool bfs(int s,int g,int cost) {
    int dist[100100];
    queue<P> que;
    if(mp[{min(s,g),max(s,g)}]) {
        dist[g]=mp[{min(s,g),max(s,g)}];
        goto oliver;
    }
    que.push({0,s});
    fill(dist,dist+100100,0);
    while(!que.empty()) {
        P p = que.front();que.pop();
        int v = p.second;
        for(auto e:G[v]) {
            if(!dist[e.to]){
                dist[e.to]=dist[v]+e.cost;
                mp[{min(e.to,s),max(e.to,s)}]=dist[e.to];
                que.push({dist[e.to],e.to});
            }
        }
    }
    oliver:;
    if(dist[g]==cost) return true;
    // cout << g+1<< ' ' << dist[g] << endl;
    return false;
}
signed main() {
    int n,m;cin >> n >> m;
    UnionFind uf(n);
    rep(i, m) {
        int a,b,cost;cin >> a >> b >> cost;
        a--;b--;
        if(uf.same(a,b)) {
            if(!bfs(a,b,cost)) {
                printf("No\n");
                return 0;
            }
        }
        else {
            uf.unite(a,b);
            G[a].pb({b,cost});
            G[b].pb({a,-cost});
        }
    }
    printf("Yes\n");
    return 0;
}