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 long long INF = 0x3f3f3f3f3f3f3f3f;
  const long long mod = 998244353;
  inline int read() {
    int x = 0, f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (isdigit(ch)) {
      x = x * 10 + ch - 48;
      ch = getchar();
    }
    return x * f;
  }
  int mu[100007], ck[100007], prime[100007], tot;
  inline void init() {
    mu[1] = 1;
    for (int i = 2; i < 100007; i++) {
      if (!ck[i]) prime[++tot] = i, mu[i] = -1;
      for (int j = 1; j <= tot && i * prime[j] < 100007; j++) {
        int now = i * prime[j];
        ck[now] = 1;
        if (i % prime[j])
          mu[now] = -mu[i];
        else {
          mu[now] = 0;
          break;
        }
      }
    }
  }
  int t, n, m;
  struct node {
    int l, r;
  } a[57], aa[57];
  long long sum[100007], dp[55][100007];
  long long DP(int d) {
    int mm = m / d;
    for (int i = 1; i <= n; i++) {
      aa[i].l = (a[i].l - 1) / d + 1;
      aa[i].r = a[i].r / d;
    }
    sum[0] = 0;
    for (int i = 1; i <= n; i++) {
      if (aa[i].l > aa[i].r) return 0;
      for (int j = 1; j <= mm; j++) {
        if (i == 1 && j >= aa[i].l && j <= aa[i].r)
          dp[i][j] = 1;
        else if (i == 1)
          dp[i][j] = 0;
        else {
          int le = max(0, j - aa[i].r - 1), ri = max(0, j - aa[i].l);
          dp[i][j] = (sum[ri] - sum[le] + mod) % mod;
        }
      }
      for (int j = 1; j <= mm; j++) sum[j] = (sum[j - 1] + dp[i][j]) % mod;
    }
    return sum[mm];
  }
  int main() {
    t = 1;
    while (t--) {
      scanf("%d%d", &n, &m);
      init();
      for (int i = 1; i <= n; i++) {
        a[i].l = read(), a[i].r = read();
      }
      long long ans = 0;
      for (int d = 1; d <= m; d++) {
        if (!mu[d]) continue;
        ans = (ans + DP(d) * mu[d] + mod) % mod;
      }
      printf("%lld\n", ans);
    }
  }
  