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 = 0x3f3f3f3f;
  const int MAXN = 51;
  int n, m;
  int l[MAXN], r[MAXN];
  const int MAXM = 1e5 + 1;
  int mu[MAXM];
  const int MOD = 998244353;
  void add(int& A, int B) {
    A += B;
    if (A >= MOD) A -= MOD;
  }
  int dp[MAXN][100000 + 1];
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d%d", &l[i], &r[i]);
    int ans = 0;
    for (int i = 1; i <= 1e5; ++i) {
      int now = i;
      mu[i] = 1;
      for (int j = 2; j * j <= now; j++) {
        if (now % j == 0) {
          if (now % (j * j) == 0) {
            mu[i] = 0;
            break;
          }
          mu[i] *= -1;
          now /= j;
        }
      }
      if (now != 1) mu[i] *= -1;
      mu[i] = (mu[i] + MOD) % MOD;
    }
    dp[0][0] = 1;
    for (int x = 1; x <= m; ++x) {
      if (mu[x]) {
        int M = m / x;
        for (int i = 1; i <= n; ++i)
          for (int j = 1; j <= M; ++j) dp[i][j] = 0;
        for (int i = 1; i <= n; ++i) {
          int l_, r_;
          l_ = (l[i] + x - 1) / x;
          r_ = (r[i]) / x;
          if (l_ <= r_) {
            int sum = 0;
            for (int j = l_; j <= M; ++j) {
              add(sum, dp[i - 1][j - l_]);
              dp[i][j] = sum;
              if (j >= r_) add(sum, MOD - dp[i - 1][j - r_]);
            }
          }
        }
        int sum = 0;
        for (int j = 1; j <= M; ++j) add(sum, dp[n][j]);
        add(ans, 1ll * sum * mu[x] % MOD);
      }
    }
    cout << ans << endl;
    return 0;
  }
  