Problem: 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`.

Examples

Input

3 3
1 2 1
2 3 1
1 3 2


Output

Yes


Input

3 3
1 2 1
2 3 1
1 3 5


Output

No


Input

4 3
2 1 1
2 3 5
3 4 2


Output

Yes


Input

10 3
8 7 100
7 9 100
9 8 100


Output

No


Input

100 0


Output

Yes

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using pii = pair<int,int>;

const int mx = 1e5 + 5;
int vis[mx], val[mx];
int n, m;
vector<pii> G[mx];

bool dfs(int x){
    vis[x] = 1;
    for(pii u : G[x]){
        if(vis[u.first]){
            if(val[u.first] != val[x] + u.second) return false;
            else continue;
        }
        val[u.first] = val[x] + u.second;
        if(!dfs(u.first)) return false;
    }
    return true;
}

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);
    cin >> n >> m;
    for(int i = 0, a, b, w; i < m; i++){
        cin >> a >> b >> w;
        G[a].emplace_back(b,w);
        G[b].emplace_back(a,-w);
    }
    for(int i = 1; i <= n; i++){
        if(!vis[i] && !dfs(i)){
            cout << "No\n";
            return 0;
        }
    }
    cout << "Yes\n";
}