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 int64 = long long;
  const int N = 100005;
  const int Q = 998244353;
  int n, m, l[N], r[N];
  int f[2][N];
  inline void add(int &A, int B) {
    A += B;
    if (A >= Q) {
      A -= Q;
    }
  }
  bool v[N];
  int prime[N], tot;
  int mu[N];
  void init(int n) {
    int i, j, x;
    mu[1] = 1;
    for (i = 2; i <= n; ++i) {
      if (!v[i]) {
        prime[tot++] = i;
        mu[i] = -1;
      }
      for (j = 0; j < tot; ++j) {
        x = i * prime[j];
        if (x > n) break;
        v[x] = 1;
        if (i % prime[j] == 0) {
          mu[x] = 0;
          break;
        } else {
          mu[x] = -mu[i];
        }
      }
    }
  }
  void work() {
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; ++i) {
      scanf("%d%d", l + i, r + i);
    }
    init(m);
    int res = 0;
    for (int d = 1; d <= m; ++d) {
      if (mu[d] == 0) {
        continue;
      }
      int K = m / d;
      int cur = 0, nxt = 1;
      memset(f[cur], 0, sizeof(int) * (K + 1));
      f[cur][0] = 1;
      for (int i = 0; i < n; ++i) {
        memset(f[nxt], 0, sizeof(int) * (K + 1));
        int L = (l[i] + d - 1) / d;
        int R = r[i] / d;
        for (int j = 0; j <= K; ++j) {
          if (j + L <= K) {
            add(f[nxt][j + L], f[cur][j]);
          }
          if (j + R + 1 <= K) {
            add(f[nxt][j + R + 1], Q - f[cur][j]);
          }
        }
        for (int j = 1; j <= K; ++j) {
          add(f[nxt][j], f[nxt][j - 1]);
        }
        std::swap(cur, nxt);
      }
      int sum = 0;
      for (int i = 0; i <= K; ++i) {
        add(sum, f[cur][i]);
      }
      if (mu[d] > 0) {
        add(res, sum);
      } else {
        add(res, Q - sum);
      }
    }
    printf("%d\n", res);
  }
  int main() {
    int T = 1;
    while (T--) {
      work();
    }
    return 0;
  }
  