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
sys.setrecursionlimit(1000000)
from collections import deque, Counter
def getN():
    return int(input())
def getList():
    return list(map(int, input().split()))
import math
INF = 10**10
def bfs(graph, visited, position, root):
    visited[root] = 1
    position[root] = 0
    deq = deque([root])
    while(deq):
        cur = deq.pop()
        # print(cur)
        for nxt, dist in graph[cur]:
            if position[cur] + dist != position[nxt]:
                if position[nxt] != INF:
                    return False
                else:
                    position[nxt] = position[cur] + dist
            if not visited[nxt]:
                deq.append(nxt)
                visited[nxt] = 1

    return True

def main():
    n,m = getList()
    position = [INF for i in range(n)]
    graph = [[] for i in range(n)]
    for i in range(m):
        a,b,c = getList()
        graph[a-1].append((b-1, c))
        graph[b - 1].append((a-1, -c))

    # print(graph)
    visited = [0 for i in range(n)]
    for i in range(n):
        if not visited[i]:
            # print("===========")
            res = bfs(graph, visited, position, i)
            if res == False:
                print("No")
                return

    print("Yes")

if __name__ == "__main__":

    main()