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>
  const int MOD = 998244353;
  const int MAXN = 52;
  const int MAXM = 100002;
  int n, m;
  int l[MAXN], r[MAXN];
  int L[MAXN], R[MAXN];
  int mu[MAXM];
  int f[MAXN][MAXM];
  void Sieve() {
    static bool pri[MAXM];
    static std::vector<int> pml;
    mu[1] = 1;
    for (int i = 2; i < MAXM; ++i) {
      if (!pri[i]) {
        pml.push_back(i);
        mu[i] = -1;
      }
      for (auto &p : pml) {
        if (p * i >= MAXM) break;
        pri[p * i] = true;
        if (i % p == 0) {
          mu[p * i] = 0;
          break;
        }
        mu[p * i] = -mu[i];
      }
    }
  }
  int Work(int d) {
    for (int i = 1; i <= n; ++i) {
      L[i] = (l[i] + d - 1) / d;
      R[i] = r[i] / d;
      if (L[i] > R[i]) return 0;
    }
    f[0][0] = 1;
    for (int i = 1; i <= n; ++i)
      for (int j = 0; j <= m / d; ++j) f[i][j] = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = 0; j + L[i] <= m / d; ++j) {
        int st = j + L[i], ed = std::min(j + R[i] + 1, m / d + 1);
        f[i][st] = (f[i][st] + f[i - 1][j]) % MOD;
        f[i][ed] = (f[i][ed] - f[i - 1][j]) % MOD;
      }
      for (int j = 1; j <= m / d; ++j) f[i][j] = (f[i][j] + f[i][j - 1]) % MOD;
    }
    int ret = 0;
    for (int i = 0; i <= m / d; ++i) ret = (ret + f[n][i]) % MOD;
    return ret;
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) scanf("%d%d", &l[i], &r[i]);
    Sieve();
    int ans = 0;
    for (int d = 1; d <= m; ++d) ans = (ans + mu[d] * Work(d)) % MOD;
    printf("%d\n", ans < 0 ? ans + MOD : ans);
    return 0;
  }
  