#include <bits/stdc++.h>
using namespace std;
const long long N = 5e5 + 5;
const long double PI = 3.14159265358979323846;
const long double eps = 1e-11;
const long long INF = 1.5e18;
const long long inf = 1e9;
const long long M = 998244353;
const string IO[2]{"NO\n", "YES\n"};
const string Io[2]{"No\n", "Yes\n"};
const string io[2]{"no\n", "yes\n"};
const string bads = "-1\n";
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2>& a) {
  out << a.first << ' ' << a.second;
  return out;
}
template <class T>
istream& operator>>(istream& in, vector<T>& a) {
  for (auto& el : a) {
    in >> el;
  }
  return in;
}
template <class T>
ostream& operator<<(ostream& out, vector<T>& a) {
  for (auto& el : a) {
    out << el << ' ';
  }
  return out;
}
long long cnt[100005];
void solve() {
  long long n, qm;
  cin >> n >> qm;
  vector<long long> ql(n), qr(n);
  for (long long i = 0; i < n; ++i) {
    cin >> ql[i] >> qr[i];
  }
  for (long long v = 1; v <= qm; ++v) {
    bool can = 1;
    vector<long long> l(n), r(n);
    for (long long i = 0; i < n; ++i) {
      r[i] = qr[i] / v;
      l[i] = (ql[i] + v - 1) / v;
      if (l[i] > r[i]) {
        can = 0;
        break;
      }
    }
    if (!can) continue;
    long long m = qm / v;
    vector<vector<long long> > dp(n + 1, vector<long long>(m + 1, 0));
    vector<vector<long long> > sum(n + 1, vector<long long>(m + 1, 0));
    dp[0][0] = 1;
    for (long long i = 0; i <= m; ++i) {
      sum[0][i] = 1;
    }
    for (long long i = 1; i <= n; ++i) {
      for (long long j = 1; j <= m; ++j) {
        if (j - l[i - 1] < 0)
          dp[i][j] = 0;
        else {
          dp[i][j] = sum[i - 1][j - l[i - 1]];
          if (j - r[i - 1] > 0) dp[i][j] -= sum[i - 1][j - r[i - 1] - 1];
          sum[i][j] = (sum[i][j - 1] + dp[i][j]) % M;
        }
      }
    }
    for (long long i = 0; i <= m; ++i) {
      cnt[v] += dp[n][i];
      cnt[v] %= M;
    }
  }
  long long ans = cnt[1];
  vector<long long> use(qm + 1, 0);
  for (long long i = 2; i <= qm; ++i) {
    long long now = M + 1 - use[i];
    now %= M;
    ans += M - (now * cnt[i]) % M;
    ans %= M;
    for (long long j = 0; j <= qm; j += i) {
      use[j] += now;
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Please create a solution in CPP to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 5e5 + 5;
const long double PI = 3.14159265358979323846;
const long double eps = 1e-11;
const long long INF = 1.5e18;
const long long inf = 1e9;
const long long M = 998244353;
const string IO[2]{"NO\n", "YES\n"};
const string Io[2]{"No\n", "Yes\n"};
const string io[2]{"no\n", "yes\n"};
const string bads = "-1\n";
template <class T1, class T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2>& a) {
  out << a.first << ' ' << a.second;
  return out;
}
template <class T>
istream& operator>>(istream& in, vector<T>& a) {
  for (auto& el : a) {
    in >> el;
  }
  return in;
}
template <class T>
ostream& operator<<(ostream& out, vector<T>& a) {
  for (auto& el : a) {
    out << el << ' ';
  }
  return out;
}
long long cnt[100005];
void solve() {
  long long n, qm;
  cin >> n >> qm;
  vector<long long> ql(n), qr(n);
  for (long long i = 0; i < n; ++i) {
    cin >> ql[i] >> qr[i];
  }
  for (long long v = 1; v <= qm; ++v) {
    bool can = 1;
    vector<long long> l(n), r(n);
    for (long long i = 0; i < n; ++i) {
      r[i] = qr[i] / v;
      l[i] = (ql[i] + v - 1) / v;
      if (l[i] > r[i]) {
        can = 0;
        break;
      }
    }
    if (!can) continue;
    long long m = qm / v;
    vector<vector<long long> > dp(n + 1, vector<long long>(m + 1, 0));
    vector<vector<long long> > sum(n + 1, vector<long long>(m + 1, 0));
    dp[0][0] = 1;
    for (long long i = 0; i <= m; ++i) {
      sum[0][i] = 1;
    }
    for (long long i = 1; i <= n; ++i) {
      for (long long j = 1; j <= m; ++j) {
        if (j - l[i - 1] < 0)
          dp[i][j] = 0;
        else {
          dp[i][j] = sum[i - 1][j - l[i - 1]];
          if (j - r[i - 1] > 0) dp[i][j] -= sum[i - 1][j - r[i - 1] - 1];
          sum[i][j] = (sum[i][j - 1] + dp[i][j]) % M;
        }
      }
    }
    for (long long i = 0; i <= m; ++i) {
      cnt[v] += dp[n][i];
      cnt[v] %= M;
    }
  }
  long long ans = cnt[1];
  vector<long long> use(qm + 1, 0);
  for (long long i = 2; i <= qm; ++i) {
    long long now = M + 1 - use[i];
    now %= M;
    ans += M - (now * cnt[i]) % M;
    ans %= M;
    for (long long j = 0; j <= qm; j += i) {
      use[j] += now;
    }
  }
  cout << ans << '\n';
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  while (t--) {
    solve();
  }
}

```