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 long long N = 5e5 + 5;
  const long double PI = 3.14159265358979323846;
  const long double eps = 1e-11;
  const long long INF = 1.5e18;
  const long long inf = 1e9;
  const long long M = 998244353;
  const string IO[2]{"NO\n", "YES\n"};
  const string Io[2]{"No\n", "Yes\n"};
  const string io[2]{"no\n", "yes\n"};
  const string bads = "-1\n";
  template <class T1, class T2>
  istream& operator>>(istream& in, pair<T1, T2>& a) {
    in >> a.first >> a.second;
    return in;
  }
  template <class T1, class T2>
  ostream& operator<<(ostream& out, pair<T1, T2>& a) {
    out << a.first << ' ' << a.second;
    return out;
  }
  template <class T>
  istream& operator>>(istream& in, vector<T>& a) {
    for (auto& el : a) {
      in >> el;
    }
    return in;
  }
  template <class T>
  ostream& operator<<(ostream& out, vector<T>& a) {
    for (auto& el : a) {
      out << el << ' ';
    }
    return out;
  }
  long long cnt[100005];
  void solve() {
    long long n, qm;
    cin >> n >> qm;
    vector<long long> ql(n), qr(n);
    for (long long i = 0; i < n; ++i) {
      cin >> ql[i] >> qr[i];
    }
    for (long long v = 1; v <= qm; ++v) {
      bool can = 1;
      vector<long long> l(n), r(n);
      for (long long i = 0; i < n; ++i) {
        r[i] = qr[i] / v;
        l[i] = (ql[i] + v - 1) / v;
        if (l[i] > r[i]) {
          can = 0;
          break;
        }
      }
      if (!can) continue;
      long long m = qm / v;
      vector<vector<long long> > dp(n + 1, vector<long long>(m + 1, 0));
      vector<vector<long long> > sum(n + 1, vector<long long>(m + 1, 0));
      dp[0][0] = 1;
      for (long long i = 0; i <= m; ++i) {
        sum[0][i] = 1;
      }
      for (long long i = 1; i <= n; ++i) {
        for (long long j = 1; j <= m; ++j) {
          if (j - l[i - 1] < 0)
            dp[i][j] = 0;
          else {
            dp[i][j] = sum[i - 1][j - l[i - 1]];
            if (j - r[i - 1] > 0) dp[i][j] -= sum[i - 1][j - r[i - 1] - 1];
            sum[i][j] = (sum[i][j - 1] + dp[i][j]) % M;
          }
        }
      }
      for (long long i = 0; i <= m; ++i) {
        cnt[v] += dp[n][i];
        cnt[v] %= M;
      }
    }
    long long ans = cnt[1];
    vector<long long> use(qm + 1, 0);
    for (long long i = 2; i <= qm; ++i) {
      long long now = M + 1 - use[i];
      now %= M;
      ans += M - (now * cnt[i]) % M;
      ans %= M;
      for (long long j = 0; j <= qm; j += i) {
        use[j] += now;
      }
    }
    cout << ans << '\n';
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    long long t = 1;
    while (t--) {
      solve();
    }
  }
  