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 math
  import sys
  input = sys.stdin.readline
  
  n, m = map(int, input().split())
  mod = 998244353
  lr = [list(map(int, input().split())) for _ in range(n)]
  ans = 0
  s = [0] * (m + 1)
  s[1] = 1
  for j in range(1, m + 1):
      m0 = m // j
      dpsum = [1] * (m0 + 2)
      dpsum[0] = 0
      dp = [0] * (m0 + 1)
      ok = 1
      s0 = 0
      for l, r in lr:
          l0, r0 = int(math.ceil(l / j)), r // j
          if l0 > r0:
              ok = 0
              break
          l0 -= 1
          for i in range(m0 + 1):
              dp[i] = dpsum[max(0, i - l0)] - dpsum[max(0, i - r0)]
              dp[i] %= mod
          for i in range(m0 + 1):
              dpsum[i + 1] = dp[i] + dpsum[i]
              dpsum[i + 1] %= mod
      if ok:
          sj = s[j]
          ss = 0
          for i in range(m0 + 1):
              ss += dp[i]
              ss %= mod
          ans += sj * ss % mod
          ans %= mod
          for k in range(2 * j, m + 1, j):
              s[k] -= sj
  print(ans)