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 <bits/stdc++.h>
#define rep(i,n) for(int i=0; i<(n); i++)
#define FOR(i,a,n) for(int i=(a); i<(n); i++)
#define all(v) v.begin(), v.end()
#define fi first
#define se second
#define sz(x) int(x.size())
using namespace std;
using ll = long long;
using P = pair<int ,int>;
const int INF = 1001001001;
const int MOD = 1000000007;
template<class T> inline bool chmax(T &a, T b) { if (a<b) {a=b; return 1;} return 0;}
template<class T> inline bool chmin(T &a, T b) { if (b<a) {a=b; return 1;} return 0;}

#ifdef _DEBUG
#include "debug.h"
#else
#define debug(...)
#define print(...)
#endif

vector<vector<P>> G;
vector<int> loc;
vector<bool> seen;
bool dfs(int v) {
    if (loc[v]==INF) loc[v] = 0;
    seen[v] = true;
    for(const auto& p: G[v]) {
        int nv = p.fi, tmp = loc[v]+p.se;
        if (seen[nv] && loc[nv]!=tmp) return false;
        if (!seen[nv]) {
            loc[nv] = tmp;
            dfs(nv);
        }
    }
    return true;
}

int main() {
    int n, m; cin >> n >> m;
    G.resize(n);
    rep(i, m) {
        int u, v, cost; cin >> u >> v >> cost;
        u--; v--;
        G[u].emplace_back(v, cost);
        G[v].emplace_back(u, -cost);
    }
    loc.assign(n, INF);
    seen.assign(n, false);
    rep(i, n) {
        if (seen[i]) continue;
        if (!dfs(i)) {
            cout << "No" << endl;
            return 0;
        }
        debug(seen);
        debug(loc);
    }
    cout << "Yes" << endl;
    return 0;
}