#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
long long n, m, l[60], r[60], memo[200010], L[60], R[60];
long long dp(long long k) {
  if (memo[k] != -1) return memo[k];
  long long ans = 0;
  for (long long i = 2; i * k <= m; i++) {
    ans = (ans - dp(i * k) + mod) % mod;
  }
  long long M = m / k, esq = 0;
  for (long long i = 1; i <= n; i++) {
    L[i] = l[i] / k;
    if (l[i] % k) L[i]++;
    R[i] = r[i] / k;
    esq += L[i];
    if (R[i] < L[i]) return memo[k] = 0;
    R[i] -= L[i];
  }
  if (M < esq) return memo[k] = 0;
  M -= esq;
  vector<long long> soma(M + 10);
  for (long long i = 0; i <= min(R[1], M); i++) soma[i] = 1;
  for (long long obj = 2; obj <= n; obj++) {
    vector<long long> pref(M + 10), aux(M + 10);
    for (long long i = 0; i <= M; i++) {
      if (i != 0) pref[i] = pref[i - 1];
      aux[i] = soma[i];
      pref[i] = (pref[i] + soma[i]) % mod;
    }
    for (long long num = 0; num <= M; num++) {
      long long left = max(0ll, num - R[obj]);
      soma[num] = pref[num];
      if (left > 0) soma[num] = (soma[num] - pref[left - 1] + mod) % mod;
      ;
    }
  }
  for (long long i = 0; i <= M; i++) ans = (ans + soma[i]) % mod;
  return memo[k] = ans;
}
void solve() {
  memset(memo, -1, sizeof(memo));
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
  }
  cout << dp(1) << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  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 n, m, l[60], r[60], memo[200010], L[60], R[60];
long long dp(long long k) {
  if (memo[k] != -1) return memo[k];
  long long ans = 0;
  for (long long i = 2; i * k <= m; i++) {
    ans = (ans - dp(i * k) + mod) % mod;
  }
  long long M = m / k, esq = 0;
  for (long long i = 1; i <= n; i++) {
    L[i] = l[i] / k;
    if (l[i] % k) L[i]++;
    R[i] = r[i] / k;
    esq += L[i];
    if (R[i] < L[i]) return memo[k] = 0;
    R[i] -= L[i];
  }
  if (M < esq) return memo[k] = 0;
  M -= esq;
  vector<long long> soma(M + 10);
  for (long long i = 0; i <= min(R[1], M); i++) soma[i] = 1;
  for (long long obj = 2; obj <= n; obj++) {
    vector<long long> pref(M + 10), aux(M + 10);
    for (long long i = 0; i <= M; i++) {
      if (i != 0) pref[i] = pref[i - 1];
      aux[i] = soma[i];
      pref[i] = (pref[i] + soma[i]) % mod;
    }
    for (long long num = 0; num <= M; num++) {
      long long left = max(0ll, num - R[obj]);
      soma[num] = pref[num];
      if (left > 0) soma[num] = (soma[num] - pref[left - 1] + mod) % mod;
      ;
    }
  }
  for (long long i = 0; i <= M; i++) ans = (ans + soma[i]) % mod;
  return memo[k] = ans;
}
void solve() {
  memset(memo, -1, sizeof(memo));
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
  }
  cout << dp(1) << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
}

```