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:
def find(x):
    '''
    xの根を求める
    '''
    if par[x] < 0:
        return x
    else:
        p = find(par[x])
        diff_w[x] += diff_w[par[x]]
        par[x] = p
        return p


def weight(x):
    '''
    xの根からの距離を求める
    '''
    find(x)
    return diff_w[x]


def union(x, y, w):
    '''
    w[y] = w[x] + w となるようにxとyの属する集合を併合する
    '''
    w += diff_w[x] - diff_w[y]
    x = find(x)
    y = find(y)
    
    if x == y:
        return False

    if par[x] > par[y]:
        x, y = y, x
        w = -w
    par[x] += par[y]
    par[y] = x
    diff_w[y] = w
    return True


def same(x, y):
    '''
    xとyが同じ集合に属するかを判定する
    '''
    return find(x) == find(y)


def size(x):
    '''
    xが属する集合の個数を求める
    '''
    return -par[find(x)]


def diff(x, y):
    '''
    xとyが同じ集合に属するときの w[y] - w[x] を求める
    '''
    return weight(y) - weight(x)


n, m = map(int, input().split())

par = [-1] * n
diff_w = [0] * n

for _ in range(m):
    l, r, d = map(int, input().split())

    if same(l-1, r-1):
        if diff(l-1, r-1) != d:
            print('No')
            exit()
    else:
        union(l-1, r-1, d)

print('Yes')