#include <bits/stdc++.h>
using namespace std;
long long n, A[71], B[71], L[71], R[71], C[71], mm, s, DPF[100071],
    DP[71][100071];
void solve(long long t) {
  long long m = mm / t;
  s = 0;
  for (long long i = 1; i <= n; ++i) {
    L[i] = (A[i] - 1) / t + 1;
    R[i] = B[i] / t;
    s += L[i];
    C[i] = R[i] - L[i];
  }
  m -= s;
  if (m < 0) return;
  for (long long j = 0; j <= m; ++j) DP[0][j] = 1;
  for (long long i = 1; i <= n; ++i) {
    for (long long j = 0; j <= m; ++j) {
      if (j - C[i] - 1 >= 0)
        DP[i][j] = (DP[i - 1][j] - DP[i - 1][j - C[i] - 1]) % 998244353;
      else
        DP[i][j] = DP[i - 1][j];
      if (j) DP[i][j] = (DP[i][j] + DP[i][j - 1]) % 998244353;
    }
  }
  if (!DP[n][m]) {
    DPF[t] = 0;
    return;
  }
  DPF[t] = (DP[n][m] + 998244353) % 998244353;
}
signed main() {
  cin >> n >> mm;
  for (long long i = 1; i <= n; ++i) cin >> A[i] >> B[i];
  for (long long i = 1; i <= mm; ++i) solve(i);
  for (long long i = mm; i >= 1; --i) {
    for (long long j = 2 * i; j <= mm; j += i)
      DPF[i] = (DPF[i] - DPF[j] + 998244353) % 998244353;
  }
  cout << DPF[1] << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long n, A[71], B[71], L[71], R[71], C[71], mm, s, DPF[100071],
    DP[71][100071];
void solve(long long t) {
  long long m = mm / t;
  s = 0;
  for (long long i = 1; i <= n; ++i) {
    L[i] = (A[i] - 1) / t + 1;
    R[i] = B[i] / t;
    s += L[i];
    C[i] = R[i] - L[i];
  }
  m -= s;
  if (m < 0) return;
  for (long long j = 0; j <= m; ++j) DP[0][j] = 1;
  for (long long i = 1; i <= n; ++i) {
    for (long long j = 0; j <= m; ++j) {
      if (j - C[i] - 1 >= 0)
        DP[i][j] = (DP[i - 1][j] - DP[i - 1][j - C[i] - 1]) % 998244353;
      else
        DP[i][j] = DP[i - 1][j];
      if (j) DP[i][j] = (DP[i][j] + DP[i][j - 1]) % 998244353;
    }
  }
  if (!DP[n][m]) {
    DPF[t] = 0;
    return;
  }
  DPF[t] = (DP[n][m] + 998244353) % 998244353;
}
signed main() {
  cin >> n >> mm;
  for (long long i = 1; i <= n; ++i) cin >> A[i] >> B[i];
  for (long long i = 1; i <= mm; ++i) solve(i);
  for (long long i = mm; i >= 1; --i) {
    for (long long j = 2 * i; j <= mm; j += i)
      DPF[i] = (DPF[i] - DPF[j] + 998244353) % 998244353;
  }
  cout << DPF[1] << "\n";
  return 0;
}

```