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 maxn = 55;
  const int maxm = 1e5 + 5;
  const int mod = 998244353;
  int l[maxn], r[maxn], prim[maxm], tot, flag[maxm], mu[maxm], n;
  int dp[maxn][maxm], sum[maxn];
  void init(int n) {
    mu[1] = 1;
    for (int i = 2; i <= n; ++i) {
      if (!flag[i]) prim[tot++] = i, mu[i] = -1;
      for (int j = 0; j < tot && prim[j] * i <= n; ++j) {
        flag[i * prim[j]] = 1;
        if (i % prim[j] == 0) {
          mu[i * prim[j]] = 0;
          break;
        }
        mu[i * prim[j]] = -mu[i];
      }
    }
  }
  int f(int d, int m) {
    dp[0][0] = 1;
    for (int i = 0; i <= m; ++i) sum[i] = 1;
    for (int i = 1; i <= n; ++i) {
      int ls = (l[i] + d - 1) / d, rs = r[i] / d;
      for (int j = 0; j < ls; ++j) dp[i][j] = 0;
      for (int j = ls; j <= m; ++j) {
        int L = j - rs, R = j - ls;
        L = max(0, L);
        if (!L)
          dp[i][j] = sum[R];
        else
          dp[i][j] = (sum[R] - sum[L - 1]) % mod;
      }
      sum[0] = dp[i][0];
      for (int j = 1; j <= m; ++j) {
        sum[j] = (sum[j - 1] + dp[i][j]) % mod;
      }
    }
    int ans = 0;
    for (int i = 0; i <= m; ++i) {
      ans = (ans + dp[n][i]) % mod;
    }
    return ans;
  }
  int main() {
    int m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
      scanf("%d%d", &l[i], &r[i]);
    }
    init(m);
    int ans = 0;
    for (int i = 1; i <= m; ++i) {
      ans = (ans + 1ll * mu[i] * f(i, m / i) % mod) % mod;
    }
    ans = (ans + mod) % mod;
    printf("%d\n", ans);
  }
  