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 P = pair<ll, ll>;
  using PP = pair<ll, P>;
  const ll n_ = 1e5 + 100, inf = 1e18, mod = 998244353, sqrtN = 333, p = 27,
           m_ = 55;
  ll dy[4] = {-1, 0, 1, 0}, dx[4] = {0, 1, 0, -1};
  ll n, m, k, tc = 1, a, b, c, d, sum, x, y, z, w, base, ans;
  ll gcd(ll x, ll y) {
    if (!y) return x;
    return gcd(y, x % y);
  }
  ll mo[n_], L[n_], R[n_], dp[n_], pre[n_];
  void init_mobius() {
    fill(mo, mo + n_, 1);
    mo[1] = 1;
    for (ll i = 2; i * i < n_; i++) {
      if (mo[i] != 1) continue;
      for (ll j = i; j < n_; j += i) mo[j] *= -i;
      for (ll j = i * i; j < n_; j += i * i) mo[j] = 0;
    }
    for (ll i = 2; i < n_; i++) {
      if (mo[i] == i)
        mo[i] = 1;
      else if (mo[i] == -i)
        mo[i] = -1;
      else if (mo[i] < 0)
        mo[i] = 1;
      else if (mo[i] > 0)
        mo[i] = -1;
    }
  }
  ll f(ll x) {
    ll M = m / x;
    dp[0] = 1;
    for (int i = 1; i <= M; i++) dp[i] = 0;
    for (int i = 1; i <= n; i++) {
      int l = (L[i] + x - 1) / x, r = R[i] / x;
      if (l > r) return 0;
      pre[0] = dp[0];
      for (int j = 1; j <= M; j++) pre[j] = pre[j - 1] + dp[j], pre[j] %= mod;
      for (int j = 0; j <= M; j++) {
        if (j - l >= 0)
          dp[j] = pre[j - l];
        else
          dp[j] = 0;
        if (j - r - 1 >= 0) dp[j] -= pre[j - r - 1];
        dp[j] += mod;
        dp[j] %= mod;
      }
    }
    ll ret = 0;
    for (int i = 1; i <= M; i++) ret += dp[i], ret %= mod;
    return ret;
  }
  void solve() {
    ans = 0;
    init_mobius();
    cin >> n >> m;
    for (int i = 1; i <= n; i++) cin >> L[i] >> R[i];
    for (int i = 1; i <= m; i++) ans += mo[i] * f(i), ans %= mod;
    cout << (ans + mod) % mod;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    while (tc--) solve();
  }
  