#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> l(n);
  vector<int> r(n);
  for (int i = 0; i < n; ++i) {
    cin >> l[i] >> r[i];
  }
  vector<long long> ans(m + 1);
  for (int d = 1; d <= m; ++d) {
    vector<vector<long long>> dp(n + 1, vector<long long>(m / d + 1));
    vector<long long> pre(m / d + 1);
    dp[0][0] = 1;
    for (int i = 0; i < n; ++i) {
      int nl = (l[i] + d - 1) / d;
      int nr = r[i] / d;
      assert(nl > 0);
      if (nl > nr) {
        continue;
      }
      pre[0] = dp[i][0];
      for (int j = 1; j <= m / d; ++j) {
        pre[j] = (pre[j - 1] + dp[i][j]) % mod;
      }
      for (int j = nl; j <= m / d; ++j) {
        dp[i + 1][j] += pre[j - nl];
        dp[i + 1][j] += mod - (j - nr - 1 >= 0 ? pre[j - nr - 1] : 0);
        dp[i + 1][j] %= mod;
      }
    }
    ans[d] = accumulate(dp[n].begin(), dp[n].end(), 0ll) % mod;
  }
  for (int i = m; i >= 1; --i) {
    for (int j = i + i; j <= m; j += i) {
      ans[i] += mod - ans[j];
      ans[i] %= mod;
    }
  }
  cout << ans[1] << '\n';
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 int mod = 998244353;
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> l(n);
  vector<int> r(n);
  for (int i = 0; i < n; ++i) {
    cin >> l[i] >> r[i];
  }
  vector<long long> ans(m + 1);
  for (int d = 1; d <= m; ++d) {
    vector<vector<long long>> dp(n + 1, vector<long long>(m / d + 1));
    vector<long long> pre(m / d + 1);
    dp[0][0] = 1;
    for (int i = 0; i < n; ++i) {
      int nl = (l[i] + d - 1) / d;
      int nr = r[i] / d;
      assert(nl > 0);
      if (nl > nr) {
        continue;
      }
      pre[0] = dp[i][0];
      for (int j = 1; j <= m / d; ++j) {
        pre[j] = (pre[j - 1] + dp[i][j]) % mod;
      }
      for (int j = nl; j <= m / d; ++j) {
        dp[i + 1][j] += pre[j - nl];
        dp[i + 1][j] += mod - (j - nr - 1 >= 0 ? pre[j - nr - 1] : 0);
        dp[i + 1][j] %= mod;
      }
    }
    ans[d] = accumulate(dp[n].begin(), dp[n].end(), 0ll) % mod;
  }
  for (int i = m; i >= 1; --i) {
    for (int j = i + i; j <= m; j += i) {
      ans[i] += mod - ans[j];
      ans[i] %= mod;
    }
  }
  cout << ans[1] << '\n';
}

```