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>

using namespace std;

const int N = 500005;

int n, m, a[N], b[N], c[N];
bool vis[N];
long long value[N];
vector<pair<int, int> > V[N];

long long mn = 1e18;
long long mx = -1e18;

void DFS(int w) {
    mn = min(mn, value[w]);
    mx = max(mx, value[w]);
    vis[w] = true;
    for (int i = 0; i < (int)V[w].size(); i++) {
        int u = V[w][i].first;
        int weight = V[w][i].second;
        if (!vis[u]) {
            value[u] = value[w] + weight;
            DFS(u);
        }
    }
}

int main() {
    
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= m; i++) {
        scanf("%d %d %d", &a[i], &b[i], &c[i]);
        V[a[i]].push_back({b[i], c[i]});
        V[b[i]].push_back({a[i], -c[i]});
    }
    
    for (int i = 1; i <= n; i++) {
        if (!vis[i]) {
            mn = 1e18;
            mx = -1e18;
            DFS(i);
            if (mx - mn > 1e9) {
                printf("No\n");
                return 0;
            }
        }
    }
    
    for (int i = 1; i <= m; i++) {
        if (value[b[i]] - value[a[i]] != c[i]) {
            printf("No\n");
            return 0;
        }
    }
    
    printf("Yes\n");
    
    return 0;
}