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: Accepted)

//
//   insert code here...cpp
//  AtCoder Regular Contest 090
//
//  Created by Huang,Zhiqiang on 2018/2/11.
//  Copyright © 2018年 Huang,Zhiqiang. All rights reserved.
//

#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
using namespace std;

typedef long long ll;

vector<pair<int,int>> ne[100005];
ll pos[100005] = { 0 };
bool visited[100005] = { false };
int N, M;
bool valid = true;

void dfs(int node, ll x) {
    if (!valid) {
        return;
    }
    visited[node] = true;
    pos[node] = x;
    for (int i = 0; i < ne[node].size(); i++) {
        int next = ne[node][i].first;
        int step = ne[node][i].second;
        if (visited[next]) {
            if (pos[next] - pos[node] != step) {
                valid = false;
                return;
            }
        } else {
            dfs(next, pos[node] + step);
        }
    }
}

int main() {
    cin >> N >> M;
    for (int i = 0; i < M; i++) {
        int l, r, d;
        cin>> l >> r >> d;
        ne[l].push_back(make_pair(r, d));
        ne[r].push_back(make_pair(l, -d));
    }
    for (int i = 1; i <= N; i++) {
        if (!visited[i]) {
            dfs(i, 0);
        }
        if (!valid) {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;
}