#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
const int MOD = 998244353;
const ll INF = 1e18;
inline void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
ll mod_neg(ll x) {
  assert(x <= 0);
  while (x < 0) {
    x += MOD;
  }
  return x % MOD;
}
void solve() {
  ll n, m;
  cin >> n >> m;
  vector<pair<ll, ll>> arr(n);
  for (ll i = 0; i < n; i++) {
    cin >> arr[i].first >> arr[i].second;
  }
  auto ask = [&](vector<pair<ll, ll>> &v, ll m2) -> ll {
    vector<ll> ways(m2 + 1, 0);
    ways[0] = 1;
    for (ll i = 0; i < n; i++) {
      ll l = v[i].first, r = v[i].second;
      vector<ll> ways_by_pref(m2 + 1, 0);
      for (ll j = 0; j <= m2; j++) {
        if (j + l <= m2) {
          ways_by_pref[j + l] = (ways_by_pref[j + l] + ways[j]) % MOD;
          if (j + r + 1 <= m2) {
            ways_by_pref[j + r + 1] =
                (ways_by_pref[j + r + 1] + mod_neg(-ways[j])) % MOD;
          }
        }
      }
      ways[0] = ways_by_pref[0];
      for (ll i = 1; i <= m2; i++) {
        ways[i] = (ways[i - 1] + ways_by_pref[i]) % MOD;
      }
    }
    ll ans{};
    for (ll i = 0; i <= m2; i++) {
      ans = (ans + ways[i]) % MOD;
    }
    return ans;
  };
  vector<ll> f(m + 1);
  for (ll i = 1; i <= m; i++) {
    vector<pair<ll, ll>> v(n);
    for (ll j = 0; j < n; j++) {
      v[j].first = (arr[j].first + i - 1) / i;
      v[j].second = arr[j].second / i;
    }
    ll m2 = m / i;
    f[i] = ask(v, m2);
  }
  for (ll i = m; i >= 1; i--) {
    for (ll j = 2 * i; j <= m; j += i) {
      f[i] = (f[i] - f[j] + MOD) % MOD;
    }
  }
  cout << f[1] << '\n';
}
int main() {
  fastio();
  cout << setprecision(15) << fixed;
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
const int MOD = 998244353;
const ll INF = 1e18;
inline void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '{';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << '}';
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
int rand(int l, int r) {
  uniform_int_distribution<int> uid(l, r);
  return uid(rng);
}
ll mod_neg(ll x) {
  assert(x <= 0);
  while (x < 0) {
    x += MOD;
  }
  return x % MOD;
}
void solve() {
  ll n, m;
  cin >> n >> m;
  vector<pair<ll, ll>> arr(n);
  for (ll i = 0; i < n; i++) {
    cin >> arr[i].first >> arr[i].second;
  }
  auto ask = [&](vector<pair<ll, ll>> &v, ll m2) -> ll {
    vector<ll> ways(m2 + 1, 0);
    ways[0] = 1;
    for (ll i = 0; i < n; i++) {
      ll l = v[i].first, r = v[i].second;
      vector<ll> ways_by_pref(m2 + 1, 0);
      for (ll j = 0; j <= m2; j++) {
        if (j + l <= m2) {
          ways_by_pref[j + l] = (ways_by_pref[j + l] + ways[j]) % MOD;
          if (j + r + 1 <= m2) {
            ways_by_pref[j + r + 1] =
                (ways_by_pref[j + r + 1] + mod_neg(-ways[j])) % MOD;
          }
        }
      }
      ways[0] = ways_by_pref[0];
      for (ll i = 1; i <= m2; i++) {
        ways[i] = (ways[i - 1] + ways_by_pref[i]) % MOD;
      }
    }
    ll ans{};
    for (ll i = 0; i <= m2; i++) {
      ans = (ans + ways[i]) % MOD;
    }
    return ans;
  };
  vector<ll> f(m + 1);
  for (ll i = 1; i <= m; i++) {
    vector<pair<ll, ll>> v(n);
    for (ll j = 0; j < n; j++) {
      v[j].first = (arr[j].first + i - 1) / i;
      v[j].second = arr[j].second / i;
    }
    ll m2 = m / i;
    f[i] = ask(v, m2);
  }
  for (ll i = m; i >= 1; i--) {
    for (ll j = 2 * i; j <= m; j += i) {
      f[i] = (f[i] - f[j] + MOD) % MOD;
    }
  }
  cout << f[1] << '\n';
}
int main() {
  fastio();
  cout << setprecision(15) << fixed;
  int tc = 1;
  while (tc--) {
    solve();
  }
  return 0;
}

```