#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
void solve() {
  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];
  vector<long long> result(m + 1);
  for (long long g = 1; g <= m; g++) {
    long long to = m / g;
    vector<long long> dp(to + 1, 1);
    for (long long i = 0; i < n; i++) {
      long long mn = (l[i] + g - 1) / g;
      long long mx = r[i] / g;
      if (mx < mn) {
        dp[to] = 0;
        break;
      }
      vector<long long> ndp(to + 1);
      for (long long s = mn; s <= to; s++) {
        ndp[s] =
            (ndp[s - 1] + dp[s - mn] - (s - mx - 1 >= 0 ? dp[s - mx - 1] : 0)) %
            MOD;
        if (ndp[s] < 0) ndp[s] += MOD;
      }
      dp = ndp;
    }
    result[g] = dp[to];
  }
  for (long long g = m; g; g--) {
    for (long long i = 2 * g; i <= m; i += g) {
      result[g] -= result[i];
      if (result[g] < 0) result[g] += MOD;
    }
  }
  cout << result[1];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}

### Prompt

Generate 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 MOD = 998244353;
void solve() {
  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];
  vector<long long> result(m + 1);
  for (long long g = 1; g <= m; g++) {
    long long to = m / g;
    vector<long long> dp(to + 1, 1);
    for (long long i = 0; i < n; i++) {
      long long mn = (l[i] + g - 1) / g;
      long long mx = r[i] / g;
      if (mx < mn) {
        dp[to] = 0;
        break;
      }
      vector<long long> ndp(to + 1);
      for (long long s = mn; s <= to; s++) {
        ndp[s] =
            (ndp[s - 1] + dp[s - mn] - (s - mx - 1 >= 0 ? dp[s - mx - 1] : 0)) %
            MOD;
        if (ndp[s] < 0) ndp[s] += MOD;
      }
      dp = ndp;
    }
    result[g] = dp[to];
  }
  for (long long g = m; g; g--) {
    for (long long i = 2 * g; i <= m; i += g) {
      result[g] -= result[i];
      if (result[g] < 0) result[g] += MOD;
    }
  }
  cout << result[1];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}

```