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.n = n
        self.parent = [i for i in range(n + 1)]
        self.rank = [0] * (n + 1)
        self.weights = [0] * (n + 1)  # 親への距離を管理

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

    def unite(self, x, y, d):
        rx = self.find(x)
        ry = self.find(y)
        if rx == ry:
            return
        if self.rank[rx] < self.rank[ry]:
            self.parent[rx] = ry
            self.weights[rx] = d + self.weights[y] - self.weights[x]
        else:
            self.parent[ry] = rx
            self.weights[ry] = - d + self.weights[x] - self.weights[y]
            if self.rank[rx] == self.rank[ry]:
                # rank等しい時はどちらを親にしても良いが，この場合は統合されたxのrankが1増加することに注意する
                self.rank[rx] += 1

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

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


def main():
    n, m = map(int, input().split())
    wuf = WeightedUnionFind(n)
    for _ in range(m):
        l, r, d = map(int, input().split())
        l -= 1
        r -= 1
        if not wuf.same(l, r):
            wuf.unite(l, r, d)
        else:
            if wuf.diff(l, r) != d:
                print("No")
                exit()
    print("Yes")


if __name__ == "__main__":
    main()