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)

#include <bits/stdc++.h>

#define fr first
#define sc second

using namespace std;
typedef pair<int, int> ii;
typedef long long ll;
typedef unsigned long long u64;

const int NMAX = 1e5 + 3, KEK = 1453, CMAX = 'z' - 'a' + 1, INF = 2e9;
const double EPS = 1e-5;

vector<ii> g[NMAX];
vector<char> used(NMAX);
vector<int> ans(NMAX, -INF);
bool can = 1;

void dfs(int v) {
    used[v] = 1;
    if (ans[v] == -INF) {
        ans[v] = 0;
    }
    for (ii pp : g[v]) {
        int to = pp.fr, d = pp.sc;
        if (!used[to]) {
            ans[to] = ans[v] + d;
            dfs(to);
        } else if (ans[to] != ans[v] + d) {
            can = 0;
        }
    }
}

main() {
    #ifdef HOME
        freopen("input.txt", "r", stdin);
    #else
//        freopen("newphone.in", "r", stdin);
//        freopen("newphone.out", "w", stdout);
    #endif
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
        int l, r, d;
        cin >> l >> r >> d, --l, --r;
        g[l].push_back({r, d});
        g[r].push_back({l, -d});
    }
    for (int i = 0; i < n; ++i) {
        if (!used[i]) {
            dfs(i);
        }
    }
    if (can) {
        cout << "Yes";
    } else {
        cout << "No";
    }
}