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: Time Limit Exceeded)

#include <cstdio>
#include <vector>
#include <map>

using namespace std;

vector<map<int, int> > S;
map<int, int> L;

bool check(int l, int r, int d) {
    int ls = L[l];
    int rs = L[r];
    if(ls==0 && rs==0) {
        S.emplace_back();
        L[l] = S.size();
        L[r] = S.size();
        map<int, int> & s = S.back();
        s[l] = 0;
        s[r] = d;
        return true;
    }
    if(ls==0) {
        L[l] = rs;
        map<int, int> & s = S[rs-1];
        s[l] = s[r] - d;
        return true;
    }
    if(rs==0) {
        L[r] = ls;
        map<int, int> & s = S[ls-1];
        s[r] = s[l] + d;
        return true;
    }
    if(ls==rs) {
        map<int, int> & s = S[ls-1];
        return s[r]==s[l]+d;
    }
    map<int, int> & s = S[ls-1];
    map<int, int> & s2 = S[rs-1];
    if(s2.size()<s.size()) {
        int d2 = s2[r] - s[l] - d;
        for(pair<const int, int> & v : s) {
            s2[v.first] = v.second + d2;
            L[v.first] = rs;
        }
        s.clear();
    }
    else {
        int d2 = s[l] - s2[r] + d;
        for(pair<const int, int> & v : s2) {
            s[v.first] = v.second + d2;
            L[v.first] = ls;
        }
        s2.clear();
    }
    return true;
}

int main() {
    int N, M;
    scanf("%d%d", &N, &M);
    for(int m=0; m<M; ++m) {
        int l, r, d;
        scanf("%d%d%d", &l, &r, &d);--l;--r;
        if(!check(l, r, d)) {
            printf("No\n");
            return 0;
        }
    }
    printf("Yes\n");
    return 0;
}