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 long long MOD = 998244353;
  using pii = std::pair<long long, long long>;
  using namespace std;
  const long long maxn = 55, maxrange = 1e5 + 5;
  long long n, m, l[maxn], r[maxn], dp[2][maxrange], ways[maxrange];
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (long long i = 0; i < n; i++) cin >> l[i] >> r[i];
    long long maxgcd = m / n;
    for (long long gcd = maxgcd; gcd >= 1; gcd--) {
      long long compressed_till = m / gcd;
      for (long long j = 0; j <= compressed_till; j++) dp[0][j] = 0;
      dp[0][0] = 1;
      for (long long i = 0; i < n; i++) {
        long long cur = i & 1;
        long long ne = cur ^ 1;
        for (long long j = 0; j <= compressed_till; j++) dp[ne][j] = 0;
        for (long long j = 0; j < compressed_till; j++) {
          if (j > 0) {
            dp[cur][j] += dp[cur][j - 1];
            dp[cur][j] %= MOD;
          }
          long long from = j + (l[i] + gcd - 1) / gcd;
          long long to = min(j + r[i] / gcd, compressed_till);
          if (from > to) continue;
          dp[ne][from] += dp[cur][j];
          dp[ne][from] %= MOD;
          dp[ne][to + 1] += (MOD - dp[cur][j]);
          dp[ne][to + 1] %= MOD;
        }
      }
      for (long long j = 0; j <= compressed_till; j++) {
        ways[gcd] += dp[n & 1][j];
        ways[gcd] %= MOD;
      }
      for (long long mult = 2; gcd * mult <= m; mult++) {
        ways[gcd] += (MOD - ways[gcd * mult]);
        ways[gcd] %= MOD;
      }
    }
    cout << ways[1] << "\n";
    return 0;
  }
  