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>
#define rep(i,a,b) for (int i=(a); i<=int(b); i++)
#define pb push_back
#define mp make_pair
#define v first
#define w second
using namespace std;
typedef pair<int,int> pii;

const int maxn = 200009, INF = 2e9;
int d[maxn], q[maxn], l, r, a, b, c, n, m, u;
vector<pii> e[maxn];

void bfs(int s) {
	l = r = 0; q[r++] = s; d[s] = 0;
	while (l < r) {
		u = q[l++];
		for (auto k : e[u])
			if (d[k.v] >= INF) {
				d[k.v] = d[u] + k.w;
				q[r++] = k.v;
			}
			else if (d[u] + k.w != d[k.v]) {
				puts("No"); exit(0);
			}
	}
}

int main() {
	scanf("%d%d", &n, &m);
	rep (i, 1, m) {
		scanf("%d%d%d", &a, &b, &c);
		e[a].pb(mp(b, c)); e[b].pb(mp(a, -c));
	}
	memset(d, 0x7f, sizeof d);
	rep (i, 1, n)
		if (d[i] >= INF) bfs(i);
	puts("Yes");
	return 0;
}