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;

constexpr int maxn = 100000 + 10;

vector<tuple<int, int, int>> edges;
vector<pair<int, int>> vec[maxn] = {};
int poz[maxn] = {};
int in_deg[maxn] = {};

int main(){
int n, m;
cin >> n >> m;
for(int i = 0, x, y, z; i < m; ++i){
cin >> x >> y >> z;
edges.emplace_back(x, y, z);
vec[x].emplace_back(y, z);
++in_deg[y];
}
queue<int> de_viz;
for(int i =0; i < n; ++i)
if(in_deg[i] == 0)
de_viz.push(i);

while(!de_viz.empty()){
int cur = de_viz.front();
de_viz.pop();
for(auto next : vec[cur]){
poz[next.first] = poz[cur] + next.second;
if(--in_deg[next.first] == 0)
de_viz.push(next.first);
}
}
bool good = true;
for(auto& x : edges)
good = (good && poz[get<0>(x)] + get<2>(x) == poz[get<1>(x)]);
cout << (good ? "Yes" : "No");
return 0;
}