Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes

Solution in Python (3.4.3) (judge verdict: Runtime Error)

N, M = map(int, input().split())
L = []
R = []
D = []
for i in range(M):
    l, r, d = map(int, input().split())
    L.append(l)
    R.append(r)
    D.append(d)

inf = float("inf")


class WDG():  # 重み付き有向グラフ(隣接リスト表現)
    class node():
        def __init__(self, index=0, weight=0, next=None):
            self.index = index
            self.weight = weight
            self.next = next

    def __init__(self, N, M, L, R, D):
        self.heads = []
        self.x = [None] * N
        self.N = N
        for i in range(N):
            head = WDG.node()
            head.next = head
            head.index = i
            self.heads.append(head)
        for i in range(M):
            new1 = WDG.node(R[i] - 1, D[i], head)
            #new2 = WDG.node(L[i] - 1, -D[i], head)
            self.insert(L[i] - 1, new1)
            #self.insert(R[i] - 1, new2)

    def insert(self, start, new):
        new.next = self.heads[start].next
        self.heads[start].next = new

    def show_graph(self):
        for i in self.heads:
            node = i.next
            print("{}=>".format(i.index + 1), end="")
            while node != i:
                print("{}({})".format(node.index + 1, node.weight), end=" ")
                node = node.next
            print("")

    def _DFS(self, s_head):  # 始点sから到達できる連結成分を深さ優先探索
        node = s_head.next
        while node != s_head:
            new_x = self.x[s_head.index] + node.weight
            if self.x[node.index] is not None and self.x[node.index] != new_x:
                return False
            elif self.x[node.index] == new_x:
                pass
            else:
                #print("node{}".format(node.index + 1))
                self.x[node.index] = new_x
                self._DFS(self.heads[node.index])
            node = node.next
        return True

    def DFS(self):  # すべての点を訪れるまで深さ優先探索
        for i in range(self.N):
            #print("{}個目の連結成分".format(i + 1))
            node = self.heads[i]
            if self.x[node.index] is None:
                self.x[node.index] = 0
                if not self._DFS(self.heads[node.index]):
                    return "No"
        return "Yes"


G = WDG(N, M, L, R, D)
# G.show_graph()
print(G.DFS())