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.par = [i for i in range(n)]
        self.rank = [0 for i in range(n)]
        # 親ノードとの重み
        self.diff_weight = [0 for i in range(n)]

    def find(self, x):
        if self.par[x] == x:
            return x
        # 親ノードはまだ上書きしない
        root = self.find(self.par[x])
        self.diff_weight[x] += self.diff_weight[self.par[x]]
        self.par[x] = root
        return self.par[x]

    def weight(self, x):
        """ x 根に対する重みを返す """
        self.find(x)
        return self.diff_weight[x]

    def diff(self, x, y):
        """ x と y の差分をとる """
        return self.weight(y) - self.weight(x)

    def unite(self, x, y, w):
        """ ノード x の下にノード y をつける。辺の重みは、w """
        # ただし、実際には、x の根と y の 根をくっつけるので、補正
        w += self.weight(x); w -= self.weight(y)
        x = self.find(x); y = self.find(y)
        if x == y:
            return
        if self.rank[x] < self.rank[y]:
            # y の下に x をつけることになるので、重さは反転させる
            self.par[x] = y
            w = -w
            self.diff_weight[x] = w
        elif self.rank[y] < self.rank[x]:
            self.par[y] = x
            self.diff_weight[y] = w
        else:
            self.par[y] = x
            self.rank[x] += 1
            self.diff_weight[y] = w
            
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
    # これまでに入力されたものと等しくないか確認する
    if UF.find(L) == UF.find(R):
        if UF.diff(L, R) != D:
            print('No')
            break
    else:
        UF.unite(L, R, D)
else:
    print('Yes')