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 PYTHON3):

Solution:
import sys
input = sys.stdin.readline
from collections import deque
import marshal


class Graph:
    class Edge:
        def __init__(self, to, cost):
            """
            :param to:  終点ノード
            :param cost: 辺の重み
            """
            self.to, self.cost = to, cost

    def __init__(self, n, directed=False, decrement=True, destroy=False, edges=[]):
        self.n = n
        self.directed = directed
        self.decrement = decrement
        self.destroy = destroy
        self.edges = [[] for _ in range(self.n)]
        self.dist = [None] * self.n
        self.info = [-1]*self.n
        for x, y, cost in edges:
            self.add_edge(x, y, cost)

    def add_edge(self, x, y, cost):
        if self.decrement:
            x -= 1
            y -= 1
        self.edges[x].append(self.Edge(y, cost))
        if self.directed == False:
            self.edges[y].append(self.Edge(x, cost))

    def bfs(self, start, goal=-1, time=0, save=False):
        """
        :param start: スタート地点
        :param goal: ゴール地点
        :param save: True = 前回の探索結果を保持する
        :return: （ループがあっても）最短距離。存在しなければ -1
        """
        if self.decrement:
            start -= 1
            goal -= 1
        if not save:
            self.dist = [None] * self.n
        p, t, dist = start, time, 0
        self.dist[p] = 0
        next_set = deque([(p, t, dist)])

        while next_set:
            p, t, dist = next_set.popleft()
            for edge in self.edges[p]:
                q, cost = edge.to, edge.cost
                if self.dist[q] is not None:
                    if self.dist[q] != dist + cost:
                        print("No")
                        exit()
                    continue
                self.dist[q] = dist + cost
                next_set.append((q, t + 1, dist + cost))
        return True

##################################################################################################

N, M = map(int, input().split())
graph = Graph(N, directed=True, decrement=True, destroy=False)
for _ in range(M):
    x, y, cost = map(int, input().split())
    graph.add_edge(x, y, cost)
    graph.add_edge(y, x, -cost)
for i in range(1,N+1):
    if graph.dist[i-1] is None:
        graph.bfs(i, goal=-1, time=0, save=True)
print("Yes")