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;
  const int MOD = 998244353;
  int mul(long long x, int y) { return x * y % MOD; }
  void selfmul(int &x, int y) { x = mul(x, y); }
  int add(int x, int y) {
    if ((x += y) >= MOD) x -= MOD;
    return x;
  }
  void selfadd(int &x, int y) { x = add(x, y); }
  int sub(int x, int y) {
    if ((x -= y) < 0) x += MOD;
    return x;
  }
  int mod(int x) {
    x %= MOD;
    if (x < 0) x += MOD;
    return x;
  }
  const int N = 55;
  const int MAX = 1e5 + 1;
  vector<int> prime;
  bool isPrime[MAX];
  int mobius[MAX];
  int n, m, lef[N], rig[N];
  int f[N][MAX], res;
  void sieve() {
    memset(isPrime, true, sizeof isPrime);
    isPrime[0] = isPrime[1] = false;
    mobius[1] = 1;
    for (int i = 2; i < MAX; ++i) {
      if (isPrime[i]) {
        prime.push_back(i);
        mobius[i] = -1;
      }
      for (int j = 0; i * prime[j] < MAX; ++j) {
        isPrime[i * prime[j]] = false;
        if (i % prime[j]) {
          mobius[i * prime[j]] = mobius[i] * mobius[prime[j]];
        } else
          break;
      }
    }
  }
  int func(const vector<int> &lef, const vector<int> &rig, int lim) {
    int n = lef.size();
    for (int i = 0; i < n; ++i) {
      if (lef[i] > rig[i]) return 0;
    }
    for (int i = 0; i < n; ++i) {
      memset(f[i], 0, (lim + 1) * sizeof(int));
    }
    for (int i = lef[0]; i <= lim; ++i) {
      f[0][i] = min(i, rig[0]) - lef[0] + 1;
      selfadd(f[0][i], f[0][i - 1]);
    }
    for (int i = 1; i < n; ++i) {
      for (int j = lef[i] + 1; j <= lim; ++j) {
        f[i][j] = sub(f[i - 1][j - lef[i]], f[i - 1][j - min(j - 1, rig[i]) - 1]);
        selfadd(f[i][j], f[i][j - 1]);
      }
    }
    return f[n - 1][lim] - f[n - 1][lim - 1];
  }
  int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    sieve();
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
      cin >> lef[i] >> rig[i];
    }
    for (int d = 1; d <= m; ++d) {
      auto ceildiv = [&](int a, int b) { return a / b + bool(a % b); };
      vector<int> l, r;
      for (int i = 1; i <= n; ++i) {
        l.push_back(ceildiv(lef[i], d));
        r.push_back(rig[i] / d);
      }
      selfadd(res, mod(mobius[d] * func(l, r, m / d)));
    }
    cout << res;
    return 0;
  }
  