#include <bits/stdc++.h>
using namespace std;
const long long oo = 1e18;
const long long mod = 998244353;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> l(n), r(n);
  for (long long i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
  }
  auto modfix = [&](long long &x) {
    if (x >= mod)
      x -= mod;
    else if (x < 0)
      x += mod;
  };
  vector<long long> res(m + 1, 0);
  for (long long g = 1; g <= m; g++) {
    long long mm = m / g;
    vector<vector<long long>> pd(n + 1, vector<long long>(mm + 1, 1));
    for (long long i = n - 1; i >= 0; i--) {
      for (long long second = 1; second <= mm; second++) {
        pd[i + 1][second] += pd[i + 1][second - 1];
        modfix(pd[i + 1][second]);
      }
      for (long long second = 0; second <= mm; second++) {
        pd[i][second] = 0;
        long long a = second + ceil(1.0 * l[i] / g), b = second + r[i] / g;
        b = min(b, mm);
        if (a > b) continue;
        pd[i][second] = pd[i + 1][b] - pd[i + 1][a - 1];
        modfix(pd[i][second]);
      }
    }
    res[g] = pd[0][0];
  }
  for (long long g = m; g >= 1; g--) {
    for (long long mult = g + g; mult <= m; mult += g) {
      res[g] -= res[mult];
      modfix(res[g]);
    }
  }
  cout << res[1] << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 oo = 1e18;
const long long mod = 998244353;
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> l(n), r(n);
  for (long long i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
  }
  auto modfix = [&](long long &x) {
    if (x >= mod)
      x -= mod;
    else if (x < 0)
      x += mod;
  };
  vector<long long> res(m + 1, 0);
  for (long long g = 1; g <= m; g++) {
    long long mm = m / g;
    vector<vector<long long>> pd(n + 1, vector<long long>(mm + 1, 1));
    for (long long i = n - 1; i >= 0; i--) {
      for (long long second = 1; second <= mm; second++) {
        pd[i + 1][second] += pd[i + 1][second - 1];
        modfix(pd[i + 1][second]);
      }
      for (long long second = 0; second <= mm; second++) {
        pd[i][second] = 0;
        long long a = second + ceil(1.0 * l[i] / g), b = second + r[i] / g;
        b = min(b, mm);
        if (a > b) continue;
        pd[i][second] = pd[i + 1][b] - pd[i + 1][a - 1];
        modfix(pd[i][second]);
      }
    }
    res[g] = pd[0][0];
  }
  for (long long g = m; g >= 1; g--) {
    for (long long mult = g + g; mult <= m; mult += g) {
      res[g] -= res[mult];
      modfix(res[g]);
    }
  }
  cout << res[1] << endl;
  return 0;
}

```