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;

#define INF 1145141919810364364

int N, M;
static vector<pair<int, long long> > V[100000];
static long long Num[100000];
bool check = true;

void DFS(int p) {
	for (pair<int, long long> x : V[p]) {
		if (Num[x.first] == INF) {
			Num[x.first] = Num[p] + x.second;
			DFS(x.first);
		}
		else {
			if (Num[x.first] != Num[p] + x.second) check = false;
		}
	}
}

int main() {
	cin >> N >> M;
	for (int i = 0; i < M; i++) {
		int L, R, D;
		cin >> L >> R >> D;
		L--, R--;
		V[L].push_back({ R, D });
		V[R].push_back({ L, -D });
	}
	for (int i = 0; i < N; i++) {
		Num[i] = INF;
	}
	for (int i = 0; i < N; i++) {
		if (Num[i] == INF) {
			Num[i] = 0;
			DFS(i);
		}
	}
	if (check) cout << "Yes";
	else cout << "No";
}