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: #include <bits/stdc++.h>
  using namespace std;
  int n, m;
  int l[55], r[55];
  long long pie[100100];
  long long dp[55][100100];
  bool prime[100100];
  long long mobius[100100], nummults[100100];
  vector<long long> factors[100100];
  void comp() {
    for (int i = 1; i < 100100; i++) nummults[i] = 0, mobius[i] = 6969;
    for (int i = 1; i < 100100; i++)
      for (int j = i; j < 100100; j += i) factors[j].push_back(i);
    memset(prime, true, sizeof(prime));
    for (int i = 2; i < 100100; i++) {
      if (!prime[i]) continue;
      for (int j = i * 2; j < 100100; j += i) prime[j] = false;
    }
    mobius[1] = 1;
    mobius[2] = -1;
    for (int i = 3; i < 100100; i++) {
      if (prime[i]) {
        mobius[i] = -1;
        continue;
      }
      for (int j : factors[i]) {
        if (j == 1 || j == i) continue;
        if (!prime[j]) continue;
        int c = i / j;
        if (c % j == 0)
          mobius[i] = 0;
        else
          mobius[i] = mobius[c] * mobius[j];
        break;
      }
    }
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++) scanf("%d%d", &l[i], &r[i]);
    for (int p = 1; p <= m; p++) {
      int bound = (m / p);
      for (int i = 0; i <= bound; i++) {
        long long cv = i * p;
        if (cv >= l[0] && cv <= r[0])
          dp[0][i] = 1;
        else
          dp[0][i] = 0;
      }
      for (int j = 1; j < n; j++) {
        long long slide = 0;
        long long lv = l[j] / p;
        if (l[j] % p) lv++;
        int rv = r[j] / p;
        for (int i = 0; i <= bound; i++) {
          if (i - lv >= 0) slide += dp[j - 1][i - lv];
          if (i - rv > 0) slide -= dp[j - 1][i - rv - 1];
          slide += 998244353;
          slide %= 998244353;
          dp[j][i] = slide;
        }
      }
      pie[p] = 0;
      for (int i = 0; i <= bound; i++) pie[p] += dp[n - 1][i];
      pie[p] %= 998244353;
    }
    comp();
    long long ans = 0;
    for (int i = 1; i <= m; i++) {
      ans += mobius[i] * pie[i];
      ans += 998244353;
      ans %= 998244353;
    }
    printf("%lld\n", ans);
  }
  