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: import sys
  
  # sys.setrecursionlimit(200005)
  int1 = lambda x: int(x)-1
  p2D = lambda x: print(*x, sep="\n")
  def II(): return int(sys.stdin.readline())
  def LI(): return list(map(int, sys.stdin.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def LI1(): return list(map(int1, sys.stdin.readline().split()))
  def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
  def SI(): return sys.stdin.readline().rstrip()
  inf = 10**16
  # md = 10**9+7
  md = 998244353
  
  n, M = LI()
  lr = LLI(n)
  
  def doDP(g):
      m = M//g
      dp = [0]*(m+3)
      dp[0] = 1
      for i, (l, r) in enumerate(lr):
          l, r = (l+g-1)//g, r//g
          ndp = [0]*(m+3)
          for j in range(m):
              pre = dp[j]
              if pre == 0: continue
              nl = j+l
              if nl > m: break
              nr = min(j+r+1, m+1)
              ndp[nl] = (ndp[nl]+pre)%md
              ndp[nr] = (ndp[nr]-pre)%md
          for j in range(m):
              ndp[j+1] = (ndp[j+1]+ndp[j])%md
          dp = ndp
      res = 0
      for v in dp[1:m+1]: res = (res+v)%md
      return res
  
  ans = [0]
  for g in range(1, M+1):
      ans.append(doDP(g))
  
  for i in range(M-1, 0, -1):
      for j in range(i+i, M+1, i):
          ans[i] -= ans[j]
      ans[i] %= md
  
  print(ans[1])
  