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())

INF = 10 ** 9 + 1
G, dist = [[] for i in range(N)], [INF for j in range(N)]
check, p = [0 for k in range(N)], [-1 for l in range(N)]


def dijkstra(s):
    dist[s], pq = 0, []
    heapq.heappush(pq, (0, s))

    while len(pq) != 0:
        u = heapq.heappop(pq)
        check[u[1]] = 2

        if dist[u[1]] < u[0]:
            continue

        for v in G[u[1]]:
            if check[v[0]] != 2:
                if dist[u[1]] + v[1] < dist[v[0]]:
                    dist[v[0]], check[v[0]] = dist[u[1]] + v[1], 1
                    heapq.heappush(pq, (dist[v[0]], v[0]))


def main():
    data = []
    for i in range(M):
        l, r, d = map(int, input().split())
        G[l-1].append((r-1, d))
        p[r-1] = l - 1
        data.append((l, r, d))

    for i in range(N):
        if p[i] == -1:
            dijkstra(i)

    for i in range(M):
        l, r, d = data[i]
        if dist[r-1] - dist[l-1] != d:
            print("No")
            break
    else:
        print("Yes")


if __name__ == '__main__':
    main()