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 Ruby (2.3.3) (judge verdict: Runtime Error)


n,h=gets.chomp.split(" ").map { |e| e.to_i  }
a=Array.new(n)

n.times do |i|
  x,y = gets.chomp.split(" ").map { |e| e.to_i }
  a[i] = {"a" => x,"b" => y}
end

count = 0

def quick_sort_hash(c,x)
  if c==[]
    return []
  end
  same = c.select{ |h| h[x] == c[0][x] }
  smaller = c.select{ |h| h[x] < c[0][x] }
  larger = c.select { |h| h[x] > c[0][x] }

  return quick_sort_hash(smaller,x) + same + quick_sort_hash(larger,x)

end
  max_a = {"a" => 0,"b" => 0}
  indx = 0


for i in 0..n-1 do
  if a[i]["a"] > max_a["a"]
    max_a = a[i]
    indx = i
  end
end

a.slice!(indx,1)
a = quick_sort_hash(a,"b").reverse


i = 0

while h > 0 && i <= n-2 && a[i]["b"] > max_a["a"]
  h -= a[i]["b"]
  count += 1
  i += 1
end


if h <= 0
  puts count
elsif h <= max_a["b"]
  puts count + 1
else
  puts count + 1 + (h - max_a["b"] + max_a["a"] - 1)/max_a["a"]
end