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;
  using ll = long long;
  using ld = long double;
  using ull = unsigned long long;
  const int MOD = 998244353;
  const ll INF = 1e18;
  inline void fastio() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
  }
  template <typename A, typename B>
  ostream &operator<<(ostream &os, const pair<A, B> &p) {
    return os << '(' << p.first << ", " << p.second << ')';
  }
  template <typename T_container, typename T = typename enable_if<
                                      !is_same<T_container, string>::value,
                                      typename T_container::value_type>::type>
  ostream &operator<<(ostream &os, const T_container &v) {
    os << '{';
    string sep;
    for (const T &x : v) os << sep << x, sep = ", ";
    return os << '}';
  }
  void dbg_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void dbg_out(Head H, Tail... T) {
    cerr << ' ' << H;
    dbg_out(T...);
  }
  mt19937 rng(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  int rand(int l, int r) {
    uniform_int_distribution<int> uid(l, r);
    return uid(rng);
  }
  ll mod_neg(ll x) {
    assert(x <= 0);
    while (x < 0) {
      x += MOD;
    }
    return x % MOD;
  }
  void solve() {
    ll n, m;
    cin >> n >> m;
    vector<pair<ll, ll>> arr(n);
    for (ll i = 0; i < n; i++) {
      cin >> arr[i].first >> arr[i].second;
    }
    auto ask = [&](vector<pair<ll, ll>> &v, ll m2) -> ll {
      vector<ll> ways(m2 + 1, 0);
      ways[0] = 1;
      for (ll i = 0; i < n; i++) {
        ll l = v[i].first, r = v[i].second;
        vector<ll> ways_by_pref(m2 + 1, 0);
        for (ll j = 0; j <= m2; j++) {
          if (j + l <= m2) {
            ways_by_pref[j + l] = (ways_by_pref[j + l] + ways[j]) % MOD;
            if (j + r + 1 <= m2) {
              ways_by_pref[j + r + 1] =
                  (ways_by_pref[j + r + 1] + mod_neg(-ways[j])) % MOD;
            }
          }
        }
        ways[0] = ways_by_pref[0];
        for (ll i = 1; i <= m2; i++) {
          ways[i] = (ways[i - 1] + ways_by_pref[i]) % MOD;
        }
      }
      ll ans{};
      for (ll i = 0; i <= m2; i++) {
        ans = (ans + ways[i]) % MOD;
      }
      return ans;
    };
    vector<ll> f(m + 1);
    for (ll i = 1; i <= m; i++) {
      vector<pair<ll, ll>> v(n);
      for (ll j = 0; j < n; j++) {
        v[j].first = (arr[j].first + i - 1) / i;
        v[j].second = arr[j].second / i;
      }
      ll m2 = m / i;
      f[i] = ask(v, m2);
    }
    for (ll i = m; i >= 1; i--) {
      for (ll j = 2 * i; j <= m; j += i) {
        f[i] = (f[i] - f[j] + MOD) % MOD;
      }
    }
    cout << f[1] << '\n';
  }
  int main() {
    fastio();
    cout << setprecision(15) << fixed;
    int tc = 1;
    while (tc--) {
      solve();
    }
    return 0;
  }
  