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;
  long long int modInverse(long long int a, long long int m) {
    long long int m0 = m;
    long long int y = 0, x = 1;
    if (m == 1) return 0;
    while (a > 1) {
      long long int q = a / m;
      long long int t = m;
      m = a % m, a = t;
      t = y;
      y = x - q * y;
      x = t;
    }
    if (x < 0) x += m0;
    return x;
  }
  void solve() {
    long long int n, m;
    cin >> n >> m;
    long long int mod = 998244353;
    long long int dp[m + 1], pp[m + 1];
    for (int j = 0; j <= m; j++) {
      dp[j] = 0;
    }
    dp[0] = 1;
    long long int A[n], B[n];
    for (long long int i = 0; i < n; i++) {
      long long int l, r;
      cin >> l >> r;
      A[i] = l;
      B[i] = r;
      for (int j = 0; j <= m; j++) {
        pp[j] = 0;
      }
      long long int sum = 0;
      for (int j = l; j <= m; j++) {
        sum += dp[j - l];
        sum %= mod;
        if (j - r - 1 >= 0) {
          sum -= dp[j - r - 1];
        }
        sum %= mod;
        sum += mod;
        sum %= mod;
        pp[j] = sum;
      }
      for (int j = 0; j <= m; j++) {
        dp[j] = pp[j];
      }
    }
    for (int j = 2; j <= m; j++) {
      long long int flag = 1, val = 0;
      long long int k = j;
      for (int l = 2; l * l <= k; l++) {
        int gg = 0;
        while (k % l == 0) {
          k /= l;
          gg++;
          if (gg > 1) {
            flag = 0;
            break;
          }
        }
        val += gg;
        if (flag == 0) {
          break;
        }
      }
      if (k > 1) {
        val++;
      }
      if (flag == 0) {
        continue;
      }
      long long int m1 = m / j, l1, r1;
      long long int dp1[m1];
      long long int pp1[m1];
      for (int i = 0; i <= m1; i++) {
        dp1[i] = 0;
      }
      dp1[0] = 1;
      for (int y = 0; y < n; y++) {
        l1 = A[y];
        l1 = (l1 + j - 1) / j;
        r1 = B[y];
        r1 /= j;
        for (int i = 0; i <= m1; i++) {
          pp1[i] = 0;
        }
        long long int sum = 0;
        for (int i = l1; i <= m1; i++) {
          sum += dp1[i - l1];
          sum %= mod;
          if (i - r1 - 1 >= 0) {
            sum -= dp1[i - r1 - 1];
          }
          sum %= mod;
          sum += mod;
          sum %= mod;
          pp1[i] = sum;
        }
        for (int i = 0; i <= m1; i++) {
          dp1[i] = pp1[i];
          if (y == n - 1) {
            if (val % 2 == 1) {
              dp[i] -= dp1[i];
              dp[i] %= mod;
              dp[i] += mod;
              dp[i] %= mod;
            } else {
              dp[i] += dp1[i];
              dp[i] %= mod;
            }
          }
        }
      }
    }
    long long int ans = 0;
    for (int i = 1; i <= m; i++) {
      ans += dp[i];
      ans %= mod;
    }
    cout << ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t = 1;
    while (t--) solve();
  }
  