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:
class WeightedUnionFind():
    def __init__(self, n):
        self.parents = [-1] * n
        self.par_weight = [0] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            root = self.find(self.parents[x])
            self.par_weight[x] += self.par_weight[self.parents[x]]
            self.parents[x] = root
            return self.parents[x]

    def union(self, x, y, w):
        w = w + self.weight(x) - self.weight(y)
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            self.parents[y] += self.parents[x]
            self.parents[x] = y
            self.par_weight[x] = -w
        else:
            self.parents[x] += self.parents[y]
            self.parents[y] = x
            self.par_weight[y] = w

    def weight(self, x):
        if self.parents[x] < 0:
            return 0
        else:
            return self.par_weight[x] + self.weight(self.parents[x])

    def diff(self, x, y):
        if self.find(x) != self.find(y):
            raise Exception('"{}" belongs to a different tree from "{}"'.format(x, y))
        return self.weight(y) - self.weight(x)


N, M = map(int, input().split())
uf = WeightedUnionFind(N)
for i in range(M):
    L, R, D = map(int, input().split())
    L -= 1
    R -= 1
    if uf.find(L) == uf.find(R):
        if uf.diff(L, R) != D:
            res = 'No'
            break
    else:
        uf.union(L, R, D)
else:
    res = 'Yes'

print(res)