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 N = 100010;
  int inf = 1e9;
  int mod = 998244353;
  inline void add(int &a, int b) {
    a += b;
    if (a >= mod) a -= mod;
  }
  inline void sub(int &a, int b) {
    a -= b;
    if (a < 0) a += mod;
  }
  inline int mul(int a, int b) { return (int)((long long)a * b % mod); }
  inline int power(int a, long long b) {
    int res = 1;
    while (b > 0) {
      if (b & 1) {
        res = mul(res, a);
      }
      a = mul(a, a);
      b >>= 1;
    }
    return res;
  }
  inline int inv(int a) {
    a %= mod;
    if (a < 0) a += mod;
    int b = mod, u = 0, v = 1;
    while (a) {
      int t = b / a;
      b -= t * a;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    assert(b == 1);
    if (u < 0) u += mod;
    return u;
  }
  vector<int> dp[2][N];
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    int l[n + 1], r[n + 1];
    for (int i = 1; i <= n; i++) {
      cin >> l[i] >> r[i];
    }
    for (int i = 0; i < 2; i++) {
      for (int j = 1; j <= m; j++) {
        dp[i][j].resize(m / j + 1, 0);
      }
    }
    for (int D = m; D >= 1; D--) {
      int last = 0;
      dp[0][D][0] = 1;
      for (int i = 1; i <= m / D; i++) add(dp[0][D][i], dp[0][D][i - 1]);
      for (int i = 1; i <= n; i++) {
        last ^= 1;
        int y = (l[i] + D - 1) / D;
        int z = r[i] / D;
        if (y > z) {
          for (int j = 0; j <= m / D; j++) {
            dp[last ^ 1][D][j] = 0;
          }
          continue;
        }
        for (int j = 0; j <= m / D; j++) {
          if (j - y >= 0) add(dp[last][D][j], dp[last ^ 1][D][j - y]);
          if (j - z > 0) sub(dp[last][D][j], dp[last ^ 1][D][j - z - 1]);
        }
        for (int j = 0; j <= m / D; j++) {
          if (j) add(dp[last][D][j], dp[last][D][j - 1]);
          dp[last ^ 1][D][j] = 0;
        }
      }
    }
    vector<int> ans(m + 1);
    for (int i = 1; i <= m; i++) ans[i] = dp[n & 1][i][m / i];
    for (int i = m; i > 0; i--) {
      for (int j = i + i; j <= m; j += i) {
        sub(ans[i], ans[j]);
      }
    }
    cout << ans[1] << "\n";
    return 0;
  }
  