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 int32 = int;
  using int64 = long long;
  using i3 = int32;
  using i4 = int64;
  using pi33 = std::pair<i3, i3>;
  using pi34 = std::pair<i3, i4>;
  using pi43 = std::pair<i4, i3>;
  using pi44 = std::pair<i4, i4>;
  const int B = 2;
  const int N = 1000005;
  const int P = 998244353;
  int n, m, d, x, y, z, D, M, c;
  int l[N], r[N], p[N], vis[N], e[N];
  int64 ans, sum;
  int64 dp[N][B];
  int main() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0), std::cout.tie(0);
    std::cin >> n >> m;
    for (int i = 2; i <= m; i++) {
      if (!vis[i]) p[++p[0]] = i;
      for (int j = 1; j <= p[0] && i * p[j] <= m; j++) {
        vis[i * p[j]] = p[j];
        if (i % p[j] == 0) break;
      }
    }
    for (int i = 1; i <= n; ++i) std::cin >> l[i] >> r[i];
    e[1] = 1;
    for (int i = 1; i <= m; ++i) {
      if (!vis[i]) {
        e[i] = i == 1 ? 1 : -1;
      } else {
        if (!e[i / vis[i]] || i / vis[i] % vis[i] == 0) continue;
        e[i] = -e[i / vis[i]];
      }
      M = m / i, x = i;
      for (int i = 1; i <= n; ++i) M -= (l[i] - 1) / x;
      for (int i = 0; i <= M; ++i) dp[i][c] = 0;
      dp[0][c] = 1;
      for (int i = 1; i <= n; ++i) {
        D = r[i] / x - (l[i] - 1) / x;
        for (int k = 1; k <= M; ++k) {
          dp[k][c] = (dp[k - 1][c] + dp[k][c]) % P;
        }
        for (int k = 0; k <= M; ++k) {
          dp[k][!c] = dp[k - 1][c];
          if (k > D) dp[k][!c] -= dp[k - D - 1][c];
        }
        c ^= 1;
      }
      sum = 0;
      for (int i = 0; i <= M; ++i) sum += dp[i][c];
      ans += (sum * e[i] % P + P) % P;
    }
    std::cout << ans % P << std::endl;
    return 0;
  }
  