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;

typedef long long ll;
typedef pair <int, int> pii;

const int MAXN = 2e5 + 5;

int N, M;
vector <pii> E[MAXN];
int l[MAXN], r[MAXN], d[MAXN];
bool bio[MAXN];
ll x[MAXN];

void load() {
	scanf("%d%d", &N, &M);
	for (int i = 0; i < M; i++) {
		scanf("%d%d%d", l + i, r + i, d + i);
		E[l[i]].push_back({r[i], d[i]});
		E[r[i]].push_back({l[i], -d[i]});
	}
}

void dfs(int node, ll curr) {
	bio[node] = true;
	x[node] = curr;
	for (auto it : E[node])
		if (!bio[it.first])
			dfs(it.first, curr + it.second);
}

bool solve() {
	for (int i = 1; i <= N; i++)
		if (!bio[i])	
			dfs(i, 0);
	
	for (int i = 0; i < M; i++)
		if (x[r[i]] - x[l[i]] != d[i])
			return false;
	
	return true;
}

int main() {
	load();
	puts(solve() ? "Yes" : "No");
	return 0;
}