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 INF_MAX = 1 << 30;
  const double INF_MIN = 1e-6;
  const int MAX_N = 50 + 5;
  const int MAX_M = 100000 + 5;
  const long long int MOD = 998244353;
  const double PI = acos(-1.0);
  const double E = exp(1.0);
  long long int n, m;
  long long int l[MAX_N];
  long long int r[MAX_N];
  long long int vis[MAX_M];
  long long int prime[MAX_M];
  long long int mu[MAX_M];
  long long int dp[MAX_N][MAX_M];
  long long int dSum[MAX_N][MAX_M];
  void prepare() {
    memset(vis, 0, sizeof(vis));
    mu[1] = 1;
    int cnt = 0;
    for (int i = 2; i < MAX_M; i++) {
      if (!vis[i]) {
        prime[cnt++] = i;
        mu[i] = -1;
      }
      for (int j = 0; j < cnt && i * prime[j] < MAX_M; j++) {
        vis[i * prime[j]] = 1;
        if (i % prime[j]) {
          mu[i * prime[j]] = -mu[i];
        } else {
          mu[i * prime[j]] = 0;
          break;
        }
      }
    }
  }
  void init() {}
  void input() {
    scanf("%lld%lld", &n, &m);
    for (int i = 1; i <= n; i++) {
      scanf("%lld%lld", &l[i], &r[i]);
    }
  }
  long long int calDp(int d) {
    for (int j = 0; j <= m / d; j++) {
      dSum[0][j] = j + 1;
    }
    for (int i = 1; i <= n; i++) {
      long long int li = (l[i] % d == 0) ? l[i] / d : l[i] / d + 1, ri = r[i] / d;
      dSum[i][0] = (li == 0);
      for (int j = 1; j <= m / d; j++) {
        if (li > ri) {
          dp[i][j] = 0;
        } else {
          int s = j - ri, e = j - li;
          dp[i][j] = ((e >= 0) ? dSum[i - 1][e] : 0) -
                     ((s - 1 >= 0) ? dSum[i - 1][s - 1] : 0);
          dp[i][j] = (dp[i][j] + MOD) % MOD;
        }
        dSum[i][j] = (dSum[i][j - 1] + dp[i][j]) % MOD;
      }
    }
    return dp[n][m / d];
  }
  void solve() {
    long long int ans = 0;
    for (int d = 1; d <= m; d++) {
      long long int fd = calDp(d);
      ans = (ans + mu[d] * fd + MOD) % MOD;
    }
    printf("%lld\n", ans);
  }
  int main() {
    ios::sync_with_stdio(false);
    prepare();
    int t;
    t = 1;
    while (t--) {
      init();
      input();
      solve();
    }
    return 0;
  }
  