Problem:
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The Maximum Number of Customers

$N$ persons visited a restaurant. The restaurant is open from 0 to $T$. The $i$-th person entered the restaurant at $l_i$ and left at $r_i$. Find the maximum number of persons during the business hours.

Constraints

$ 1 \leq N \leq 10^5 $

$ 1 \leq T \leq 10^5 $

$ 0 \leq l_i < r_i \leq T $

Input

The input is given in the following format.

$N$ $T$

$l_1$ $r_1$

$l_2$ $r_2$

:

$l_N$ $r_N$

Output

Print the maximum number of persons in a line.

Sample Input 1

6 10
0 2
1 3
2 6
3 8
4 10
5 10

Sample Output 1

4

Sample Input 2

2 2
0 1
1 2

Sample Output 2

1

Solution in Ruby (judge verdict: Accepted)

def add(a, w, d)
  while a <= d.size - 1
    d[a] += w
    a += a & -a
  end
end

def sum(i, d)
  ret = 0
  x = i
  while x > 0
    ret += d[x]
    x -= x & -x
  end
  return ret
end

n = gets.split.map(&:to_i)[0]

time = {}
customer = Array.new(n){[]}
n.times{|i|
  l, r = gets.split.map(&:to_i)
  time[l] = 1
  time[r] = 1
  customer[i] = [l, r]
}
#p customer

st = time.keys.sort
#p st
ti = {}
st.each_with_index{|v, i|
  ti[v] = i+1
}
#p ti

d = Array.new(ti.size+1){0}

n.times{|i|
  add(ti[customer[i][0]], 1, d)
  add(ti[customer[i][1]], -1, d)
}

max = 0
st.size.times{|i|
  s = sum(i+1, d)
  max = s if s > max
}
p max