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 WeightedUnionFindTree:
    """Disjoint-Set Data Structure with weight

    complexity:
        - init: O(n)
        - find, unite, same: O(alpha(n))
    """

    def __init__(self, n_nodes):
        self.par = [i for i in range(n_nodes)]  # parent
        self.rank = [0] * n_nodes  # depth of tree
        self.weights = [0] * n_nodes

    def find(self, x):
        if self.par[x] == x:
            return x
        root = self.find(self.par[x])
        self.weights[x] += self.weights[self.par[x]]
        self.par[x] = root
        return root

    def unite(self, x, y, weight=0):
        root_x, root_y = self.find(x), self.find(y)
        if root_x == root_y:
            return
        if self.rank[root_x] < self.rank[root_y]:
            self.par[root_x] = root_y
            self.weights[root_x] = weight - self.weights[x] + self.weights[y]
        else:
            self.par[root_y] = root_x
            self.weights[root_y] = -weight - self.weights[y] + self.weights[x]
            if self.rank[root_x] == self.rank[root_y]:
                self.rank[root_x] += 1

    def is_same(self, x, y):
        return self.find(x) == self.find(y)

    def diff(self, x, y):
        return self.weights[x] - self.weights[y]


N, M = map(int, input().split())
uft = WeightedUnionFindTree(N)

for _ in range(M):
    l, r, d = map(int, input().split())
    l, r = l-1, r-1
    if not uft.is_same(l, r):
        uft.unite(l, r, d)
    else:
        if uft.diff(l, r) != d:
            print("No")
            quit()
else:
    print("Yes")