#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long l[55], r[55];
long long ans[100005];
void solve() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
  }
  for (long long x = m / n; x >= 1; x--) {
    long long sz = m / x;
    vector<long long> dp(sz + 1);
    dp[0] = 1;
    for (long long i = 1; i <= n; i++) {
      vector<long long> ndp(sz + 1);
      long long mini = 1e9, maxi = 0;
      for (long long num = 1; num * x <= r[i]; num++) {
        if (num * x < l[i]) continue;
        mini = min(mini, num);
        maxi = max(maxi, num);
      }
      for (long long j = 1; j <= sz; j++) {
        dp[j] = (dp[j] + dp[j - 1]) % mod;
      }
      for (long long j = 0; j <= sz; j++) {
        long long lower = j - maxi, upper = j - mini;
        if (upper < 0) continue;
        if (lower <= 0) {
          ndp[j] = (dp[upper]) % mod;
          continue;
        }
        ndp[j] = (dp[upper] - dp[lower - 1] + mod) % mod;
      }
      dp = ndp;
    }
    for (auto u : dp) ans[x] = (ans[x] + u);
    for (long long num = 2 * x; num <= m; num += x) {
      ans[x] = (ans[x] - ans[num] + mod) % mod;
    }
  }
  cout << ans[1];
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long l[55], r[55];
long long ans[100005];
void solve() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
  }
  for (long long x = m / n; x >= 1; x--) {
    long long sz = m / x;
    vector<long long> dp(sz + 1);
    dp[0] = 1;
    for (long long i = 1; i <= n; i++) {
      vector<long long> ndp(sz + 1);
      long long mini = 1e9, maxi = 0;
      for (long long num = 1; num * x <= r[i]; num++) {
        if (num * x < l[i]) continue;
        mini = min(mini, num);
        maxi = max(maxi, num);
      }
      for (long long j = 1; j <= sz; j++) {
        dp[j] = (dp[j] + dp[j - 1]) % mod;
      }
      for (long long j = 0; j <= sz; j++) {
        long long lower = j - maxi, upper = j - mini;
        if (upper < 0) continue;
        if (lower <= 0) {
          ndp[j] = (dp[upper]) % mod;
          continue;
        }
        ndp[j] = (dp[upper] - dp[lower - 1] + mod) % mod;
      }
      dp = ndp;
    }
    for (auto u : dp) ans[x] = (ans[x] + u);
    for (long long num = 2 * x; num <= m; num += x) {
      ans[x] = (ans[x] - ans[num] + mod) % mod;
    }
  }
  cout << ans[1];
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t = 1;
  while (t--) {
    solve();
  }
}

```