#include <bits/stdc++.h>
using namespace std;
const long long M = 998244353;
vector<int> mu((int)1e5 + 1, 0);
vector<int> primes;
vector<int> isprime((int)1e5 + 1, 1);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  mu[1] = 1;
  for (int i = 2; i < mu.size(); i++) {
    if (isprime[i]) {
      mu[i] = -1;
      primes.push_back(i);
    }
    for (auto j : primes) {
      if (j * i >= mu.size()) break;
      isprime[i * j] = 0;
      if (i % j != 0)
        mu[i * j] = mu[i] * mu[j];
      else {
        mu[i * j] = 0;
        break;
      }
    }
  }
  long long n, m;
  cin >> n >> m;
  vector<pair<int, int>> v(n + 1);
  long long cmax = 1e9;
  for (int i = 1; i < n + 1; i++) {
    cin >> v[i].first >> v[i].second;
    cmax = min(cmax, (long long)v[i].second);
  }
  long long ans = 0;
  for (int d = 1; d < cmax + 1; d++) {
    long long mm = m / d;
    long long dp[n + 1][mm + 1];
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (int j = 1; j < mm + 1; j++) dp[0][j] = 1;
    for (int i = 1; i < n + 1; i++) {
      long long l = (v[i].first + d - 1) / d;
      long long r = (v[i].second) / d;
      for (int j = 1; j < mm + 1; j++) {
        if (j - r - 1 >= 0) dp[i][j] -= dp[i - 1][j - r - 1];
        if (j - l >= 0) dp[i][j] += dp[i - 1][j - l];
        if (j > 1) dp[i][j] += dp[i][j - 1];
        if (dp[i][j] < 0) {
          dp[i][j] += M;
          assert(dp[i][j] > 0);
        }
        dp[i][j] %= M;
      }
    }
    ans += ((long long)mu[d] * dp[n][mm]) % M;
    ans %= M;
  }
  if (ans < 0) ans += M;
  cout << ans;
}

### Prompt

Please formulate a cpp solution 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 M = 998244353;
vector<int> mu((int)1e5 + 1, 0);
vector<int> primes;
vector<int> isprime((int)1e5 + 1, 1);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  mu[1] = 1;
  for (int i = 2; i < mu.size(); i++) {
    if (isprime[i]) {
      mu[i] = -1;
      primes.push_back(i);
    }
    for (auto j : primes) {
      if (j * i >= mu.size()) break;
      isprime[i * j] = 0;
      if (i % j != 0)
        mu[i * j] = mu[i] * mu[j];
      else {
        mu[i * j] = 0;
        break;
      }
    }
  }
  long long n, m;
  cin >> n >> m;
  vector<pair<int, int>> v(n + 1);
  long long cmax = 1e9;
  for (int i = 1; i < n + 1; i++) {
    cin >> v[i].first >> v[i].second;
    cmax = min(cmax, (long long)v[i].second);
  }
  long long ans = 0;
  for (int d = 1; d < cmax + 1; d++) {
    long long mm = m / d;
    long long dp[n + 1][mm + 1];
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for (int j = 1; j < mm + 1; j++) dp[0][j] = 1;
    for (int i = 1; i < n + 1; i++) {
      long long l = (v[i].first + d - 1) / d;
      long long r = (v[i].second) / d;
      for (int j = 1; j < mm + 1; j++) {
        if (j - r - 1 >= 0) dp[i][j] -= dp[i - 1][j - r - 1];
        if (j - l >= 0) dp[i][j] += dp[i - 1][j - l];
        if (j > 1) dp[i][j] += dp[i][j - 1];
        if (dp[i][j] < 0) {
          dp[i][j] += M;
          assert(dp[i][j] > 0);
        }
        dp[i][j] %= M;
      }
    }
    ans += ((long long)mu[d] * dp[n][mm]) % M;
    ans %= M;
  }
  if (ans < 0) ans += M;
  cout << ans;
}

```