#include <bits/stdc++.h>
using namespace std;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
int main() {
  file();
  fast();
  int n, m, mod = 998244353;
  cin >> n >> m;
  vector<pair<int, int> > v1(n);
  for (auto& it : v1) cin >> it.first >> it.second;
  auto get_div = [&](pair<int, int> p, int num) {
    return p.second / num - ((p.first - 1) / num);
  };
  auto solve = [&](int g, int sum) -> int {
    vector<pair<int, int> > tmp;
    for (auto& it : v1) {
      if (!get_div(it, g)) return 0;
      pair<int, int> x;
      x.first = (it.first - 1) / g + 1;
      x.second = x.first + get_div(it, g) - 1;
      tmp.push_back(x);
    }
    if (tmp.empty()) return 0;
    vector<int> dp(sum + 1, 1);
    for (auto& it : tmp) {
      vector<int> cur(sum + 1);
      for (int i = 1; i <= sum; i++) {
        int r = max(-1, i - it.first);
        int l = max(-1, i - it.second);
        cur[i] = (r < 0 ? 0 : dp[r]) - (l <= 0 ? 0 : dp[l - 1]);
        cur[i] = (cur[i] + mod) % mod;
      }
      for (int i = 1; i <= sum; i++) cur[i] = (cur[i] + cur[i - 1]) % mod;
      dp = cur;
    }
    return dp[sum];
  };
  vector<int> arr(m + 1, 1);
  for (int i = 1; i <= m; i++) arr[i] = solve(i, m / i);
  for (int i = m; i >= 1; i--)
    for (int j = i + i; j <= m; j += i) arr[i] = (arr[i] - arr[j] + mod) % mod;
  cout << arr[1] << '\n';
  return 0;
}

### 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;
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
int dx[] = {0, 0, 1, -1, 1, -1, 1, -1};
void file() {}
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
}
int main() {
  file();
  fast();
  int n, m, mod = 998244353;
  cin >> n >> m;
  vector<pair<int, int> > v1(n);
  for (auto& it : v1) cin >> it.first >> it.second;
  auto get_div = [&](pair<int, int> p, int num) {
    return p.second / num - ((p.first - 1) / num);
  };
  auto solve = [&](int g, int sum) -> int {
    vector<pair<int, int> > tmp;
    for (auto& it : v1) {
      if (!get_div(it, g)) return 0;
      pair<int, int> x;
      x.first = (it.first - 1) / g + 1;
      x.second = x.first + get_div(it, g) - 1;
      tmp.push_back(x);
    }
    if (tmp.empty()) return 0;
    vector<int> dp(sum + 1, 1);
    for (auto& it : tmp) {
      vector<int> cur(sum + 1);
      for (int i = 1; i <= sum; i++) {
        int r = max(-1, i - it.first);
        int l = max(-1, i - it.second);
        cur[i] = (r < 0 ? 0 : dp[r]) - (l <= 0 ? 0 : dp[l - 1]);
        cur[i] = (cur[i] + mod) % mod;
      }
      for (int i = 1; i <= sum; i++) cur[i] = (cur[i] + cur[i - 1]) % mod;
      dp = cur;
    }
    return dp[sum];
  };
  vector<int> arr(m + 1, 1);
  for (int i = 1; i <= m; i++) arr[i] = solve(i, m / i);
  for (int i = m; i >= 1; i--)
    for (int j = i + i; j <= m; j += i) arr[i] = (arr[i] - arr[j] + mod) % mod;
  cout << arr[1] << '\n';
  return 0;
}

```