description: Mocha wants to be an astrologer. There are n stars which can be seen in Zhijiang, and the brightness of the i-th star is a_i. 
  
  Mocha considers that these n stars form a constellation, and she uses (a_1,a_2,…,a_n) to show its state. A state is called mathematical if all of the following three conditions are satisfied:
  
    * For all i (1≤ i≤ n), a_i is an integer in the range [l_i, r_i].
    * ∑  _{i=1} ^ n a_i ≤ m.
    * \gcd(a_1,a_2,…,a_n)=1.
  
  
  
  Here, \gcd(a_1,a_2,…,a_n) denotes the [greatest common divisor (GCD)](https://en.wikipedia.org/wiki/Greatest_common_divisor) of integers a_1,a_2,…,a_n.
  
  Mocha is wondering how many different mathematical states of this constellation exist. Because the answer may be large, you must find it modulo 998 244 353.
  
  Two states (a_1,a_2,…,a_n) and (b_1,b_2,…,b_n) are considered different if there exists i (1≤ i≤ n) such that a_i ≠ b_i.
  
  Input
  
  The first line contains two integers n and m (2 ≤ n ≤ 50, 1 ≤ m ≤ 10^5) — the number of stars and the upper bound of the sum of the brightness of stars.
  
  Each of the next n lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ m) — the range of the brightness of the i-th star.
  
  Output
  
  Print a single integer — the number of different mathematical states of this constellation, modulo 998 244 353.
  
  Examples
  
  Input
  
  
  2 4
  1 3
  1 2
  
  
  Output
  
  
  4
  
  Input
  
  
  5 10
  1 10
  1 10
  1 10
  1 10
  1 10
  
  
  Output
  
  
  251
  
  Input
  
  
  5 100
  1 94
  1 96
  1 91
  4 96
  6 97
  
  
  Output
  
  
  47464146
  
  Note
  
  In the first example, there are 4 different mathematical states of this constellation:
  
    * a_1=1, a_2=1.
    * a_1=1, a_2=2.
    * a_1=2, a_2=1.
    * a_1=3, a_2=1.
solution: # O(n*m*log(m))
  import sys
  input = sys.stdin.buffer.readline
  
  mod = 998244353
  
  n,m = map(int,input().split())
  
  # sieve
  primes = [0]*(m+1)
  prime = [1]*(m+1)
  
  for i in range(2,m+1):
      if prime[i]:
          for j in range(2*i,m+1,i):
              prime[j] = 0
              primes[j] += 1
  
  for i in range(2,m+1):
      if prime[i]:
          for j in range(i*i,m+1,i*i):
              primes[j] = -1
  
  ranges = [list(map(int,input().split())) for i in range(n)]
  ways_gcd = [0]*(m+1)
  
  for gcd in range(1,m+1):
      converted_ranges = [[(ranges[i][0]+gcd-1)//gcd,ranges[i][1]//gcd] for i in range(n)]
      dp1 = [0]*(m//gcd + 1)
      dp1[0] = 1
      dp2 = [0]*(m//gcd + 1)
  
      for i in range(n):
          if converted_ranges[i][0] > converted_ranges[i][1]:
              dp1,dp2 = dp2,dp1
              break
  
          l,r = converted_ranges[i]
          for j in range(len(dp2)-l):
              dp2[j+l] = (dp2[j+l] + dp1[j]) % mod
              if j+r+1 < len(dp2):
                  dp2[j+r+1] = (dp2[j+r+1] - dp1[j] + mod) % mod
  
          for i in range(1,len(dp2)):
              dp2[i] = (dp2[i] + dp2[i-1]) % mod
  
          dp1,dp2 = dp2,dp1
          dp2 = [0]*len(dp2)
  
      ways_gcd[gcd] = sum(dp1) % mod
  
  
  ans = ways_gcd[1]
  
  for i in range(2,m+1):
      if primes[i] >= 2 or prime[i]:
          if prime[i] or primes[i]&1:
              ans = (ans - ways_gcd[i] + mod) % mod
          else:
              ans = (ans + ways_gcd[i]) % mod
  
  print(ans)