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>;

int x[100009];
bool vs[100009];
vector<pii> adj[100009];

void go(int now) {
    vs[now] = 1;
    for(auto& it: adj[now]) {
        if(!vs[it.first]) {
            x[it.first] = x[now] + it.second;
            go(it.first);
        }
    }
}

int main() {
    int N, M; scanf("%d%d",&N,&M);
    for(int i=0; i<M; i++) {
        int L, R, D; scanf("%d%d%d",&L,&R,&D);
        adj[L].push_back({R, D});
        adj[R].push_back({L, -D});
    }
    for(int i=1; i<=N; i++) if(!vs[i]) go(i);
    for(int i=1; i<=N; i++) {
        for(auto& it: adj[i]) {
            if(x[i] + it.second != x[it.first]) return !printf("No");
        }
    }
    puts("Yes");
    return 0;
}