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 heapq

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


visited = [False for i in range(n)]
pos = [0 for i in range(n)]

for i in range(n):
    if visited[i]:
        continue
    hq = [(i, 0)]
    while hq:
        v, p = heapq.heappop(hq)
        if visited[v]:
            if p != pos[v]:
                print("No")
                exit()
            continue
        visited[v] = True
        pos[v] = p
        for nv, d in graph[v]:
            if visited[nv]:
                if pos[nv] != p + d:
                    print('No')
                    exit()
                continue
            heapq.heappush(hq, (nv, p+d))
print('Yes')