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 <iostream>
#include <vector>
#include <cstdio>

using namespace std;

typedef pair<int, int> pii;

const int MAXN = 1e5 + 5;
int N, M;
vector<pii> neigh[MAXN];
bool tk[MAXN];
int val[MAXN];

bool Fill(int v, int col) {
	if (tk[v]) return val[v] == col;
	tk[v] = true;
	val[v] = col;
	for (int i = 0; i < neigh[v].size(); i++) {
		pii u = neigh[v][i];
		if (!Fill(u.first, val[v] + u.second)) return false;
	}
	return true;
}

int main() {
	scanf("%d %d", &N, &M);
	for (int i = 0; i < M; i++) {
		int a, b, c;
		scanf("%d %d %d", &a, &b, &c);
		neigh[a].push_back(pii(b, c));
		neigh[b].push_back(pii(a, -c));
	}
	for (int i = 1; i <= N; i++) {
		if (!tk[i]) {
			if (!Fill(i, 0)) {
				printf("No\n");
				return 0;
			}
		}
	}
	printf("Yes\n");
	return 0;
}