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:
#!/usr/bin/env python3

import sys
sys.setrecursionlimit(10 ** 6)

class Impossible(ValueError):
    pass

def main():
    n, m = map(int, input().split())
    lrds = []
    for i in range(m):
        l, r, d = map(int, input().split())
        lrds.append((l - 1, r - 1, d))

    adj = [[] for i in range(n)]
    parents = [i for i in range(n)]
    for i in range(m):
        l, r, d = lrds[i]
        union(parents, l, r)
        adj[l].append((r, d))
        adj[r].append((l, -d))

    ds = set()
    for i in range(n):
        ds.add(find(parents, i))

    pos = [None for i in range(n)]
    try:
        for i in ds:
            pos[i] = 0
            dfs(adj, pos, i)
        print("Yes")
    except Impossible:
        print("No")

def dfs(adj, pos, x):
    for y, d in adj[x]:
        if pos[y] is None:
            pos[y] = pos[x] + d
            dfs(adj, pos, y)
        elif pos[y] != pos[x] + d:
            raise Impossible

def find(p, x):
    if p[x] != x:
        p[x] = find(p, p[x])
    return p[x]

def union(p, x, y):
    p[find(p, y)] = find(p, x)


main()