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;
  constexpr int MAXM = 1e5 + 228;
  constexpr int MOD = 998244353;
  int lp[MAXM];
  array<int, MAXM> dplast;
  array<int, MAXM> dp;
  int n;
  int f(int m, vector<pair<int, int>>& lr) {
    for (int i = 0; i <= m; ++i) {
      dplast[i] = 1;
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < min(lr[i].first, m + 1); ++j) dp[j] = 0;
      for (int j = lr[i].first; j <= m; ++j) {
        dp[j] = dplast[j - lr[i].first];
        if (j - lr[i].second - 1 >= 0) {
          dp[j] -= dplast[j - lr[i].second - 1];
        }
        if (dp[j] < 0) dp[j] += MOD;
      }
      dplast[0] = dp[0];
      for (int j = 1; j <= m; ++j) {
        dplast[j] = dplast[j - 1] + dp[j];
        if (dplast[j] >= MOD) dplast[j] -= MOD;
      }
    }
    return dplast[m];
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    vector<int> sim;
    for (int i = 2; i <= MAXM; ++i) {
      if (lp[i] == 0) {
        lp[i] = i;
        sim.push_back(i);
      }
      for (int j = 0; j < sim.size() && sim[j] <= lp[i] && i * sim[j] <= MAXM;
           ++j) {
        lp[sim[j] * i] = sim[j];
      }
    }
    int m;
    cin >> n >> m;
    vector<pair<int, int>> lr(n);
    for (int i = 0; i < n; ++i) {
      cin >> lr[i].first >> lr[i].second;
    }
    int ans = 0;
    vector<pair<int, int>> lr2(n);
    for (int d = 1; d <= m; ++d) {
      int last = -1;
      int cnt = 0;
      int c = d;
      while (c > 1) {
        if (lp[c] == last) {
          cnt = -1;
          break;
        }
        last = lp[c];
        c /= last;
        ++cnt;
      }
      if (cnt == -1) continue;
      for (int i = 0; i < n; ++i) {
        lr2[i] = make_pair((lr[i].first + d - 1) / d, lr[i].second / d);
      }
      int add = f(m / d, lr2);
      if (cnt & 1) {
        ans -= add;
        if (ans < 0) ans += MOD;
      } else {
        ans += add;
        if (ans >= MOD) ans -= MOD;
      }
    }
    cout << ans;
    return 0;
  }
  