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, N = 2e5 + 5;
  int l[N], r[N], mu[N], vis[N];
  int cnt, prime[N];
  int a[N], b[N];
  void init(int n) {
    mu[1] = 1;
    for (int i = 2; i <= n; ++i) {
      if (!vis[i]) vis[i] = 1, mu[i] = mod - 1, prime[++cnt] = i;
      for (int j = 1; j <= cnt; ++j) {
        if (prime[j] > n / i) break;
        vis[i * prime[j]] = 1;
        if (i % prime[j])
          mu[i * prime[j]] = mod - mu[i];
        else
          break;
      }
    }
  }
  int f[55][100005];
  int calc(int x, int n, int m) {
    int lim = m / x;
    for (int i = 1; i <= n; ++i)
      a[i] = l[i] / x, b[i] = r[i] / x - l[i] / x, lim -= a[i];
    for (int i = 1; i <= n; ++i)
      if (b[i] <= 0) return 0;
    if (lim <= 0) return 0;
    for (int i = 1; i <= n; ++i)
      for (int j = 0; j <= lim; ++j) f[i][j] = 0;
    for (int j = 1; j <= min(b[1], lim); ++j) f[1][j] = 1;
    for (int j = 1; j <= lim; ++j) f[1][j] += f[1][j - 1];
    for (int i = 2; i <= n; ++i)
      for (int j = 1; j <= lim; ++j) {
        f[i][j] = (f[i - 1][j - 1] + f[i][j - 1]) % mod;
        if (j - b[i] - 1 >= 0)
          f[i][j] = (f[i][j] + mod - f[i - 1][j - b[i] - 1]) % mod;
      }
    return f[n][lim];
  }
  void solve() {
    int n, m;
    cin >> n >> m;
    init(n + m);
    for (int i = 1; i <= n; ++i) cin >> l[i] >> r[i], l[i]--;
    int ans = 0;
    for (int i = 1; i <= m; ++i)
      if (mu[i]) ans = (ans + 1ll * mu[i] * calc(i, n, m)) % mod;
    cout << ans << "\n";
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T = 1;
    while (T--) solve();
  }
  