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 LuaJIT (2.0.4) (judge verdict: Wrong Answer)

local ior = io.input()
local n, m = ior:read("*n", "*n")
local hasinfo = {}
local pos = {}
local parent = {}
for i = 1, n do parent[i] = i end
for i = 1, n do hasinfo[i], pos[i] = false, 0 end
local isok = true

function getroot(idx)
  while(parent[idx] ~= idx) do
    idx = parent[idx]
  end
  return idx
end

for i = 1, m do
  local l, r, d = ior:read("*n", "*n", "*n")
  if(isok) then
    local lr, rr = getroot(l), getroot(r)
    parent[l], parent[r], parent[rr] = lr, lr, lr
    if(hasinfo[l] and hasinfo[r]) then
      isok = (d == pos[r] - pos[l])
    elseif(hasinfo[l]) then
      hasinfo[r] = true
      pos[r] = pos[l] + d
    elseif(hasinfo[r]) then
      hasinfo[l] = true
      pos[l] = pos[r] - d
    else
      hasinfo[l], hasinfo[r] = true, true
      pos[l] = 0
      pos[r] = d
    end
  end
end
if(isok) then
  local parents = {}
  for i = 1, n do
    local root = getroot(i)
    if(parents[root] == nil) then
      parents[root] = {}
      parents[root].min = pos[i]
      parents[root].max = pos[i]
    else
      parents[root].min = math.min(parents[root].min, pos[i])
      parents[root].max = math.max(parents[root].max, pos[i])
    end
  end
  for p, val in pairs(parents) do
    if(1000000000 < val.max - val.min) then
      isok = false
      break
    end
  end
end
print(isok and "Yes" or "No")