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 long long mod = 998244353;
  long long sum[55][100005], f[55][100005], t[100005];
  int l[55], r[55];
  int add(int p, int x, int y) {
    if (x < 0) return 0;
    if (y < 0) return sum[p][x];
    return sum[p][x] - sum[p][y];
  }
  long long solve1(int n, int m) {
    long long ans = 0;
    for (int i = l[1]; i <= r[1]; i++) f[1][i] = 1;
    for (int i = 1; i <= m; i++) sum[1][i] = (sum[1][i - 1] + f[1][i]) % mod;
    for (int i = 2; i <= n; i++) {
      for (int j = 1; j <= m; j++) {
        f[i][j] = add(i - 1, j - l[i], j - r[i] - 1);
        sum[i][j] = (sum[i][j - 1] + f[i][j]) % mod;
      }
    }
    for (int i = 1; i <= m; i++) ans = (ans + f[n][i]) % mod;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= m; j++) f[i][j] = sum[i][j] = 0;
    return ans;
  }
  long long solve2(int n, int m, int d) {
    long long ans = 0;
    for (int i = d; i <= m; i += d)
      if (i >= l[1] && i <= r[1]) f[1][i] = 1;
    for (int i = d; i <= m; i += d) sum[1][i] = f[1][i] + sum[1][i - d];
    for (int i = 2; i <= n; i++) {
      for (int j = d; j <= m; j += d) {
        int R = j - l[i] - ((j - l[i]) % d + d) % d;
        int L = j - r[i] - ((j - r[i]) % d + d) % d;
        if ((j - r[i]) % d == 0) L -= d;
        f[i][j] = add(i - 1, R, L);
        sum[i][j] = (sum[i][j - d] + f[i][j]) % mod;
      }
    }
    for (int i = d; i <= m; i += d) ans = (ans + f[n][i]) % mod;
    for (int i = 1; i <= n; i++)
      for (int j = d; j <= m; j += d) f[i][j] = sum[i][j] = 0;
    return ans;
  }
  int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d%d", &l[i], &r[i]);
    long long ans = solve1(n, m);
    for (int d = 2; d <= m / n; d++) {
      ans = ((ans - (1 - t[d]) * solve2(n, m, d)) % mod + mod) % mod;
      for (int j = 2; j * d <= m; j++) t[j * d] += 1 - t[d];
    }
    printf("%lld", ans);
    return 0;
  }
  