#include <bits/stdc++.h>
using namespace std;
const int N = 50, M = 1e5 + 1, MOD = 998244353;
int l[N], r[N];
long long f[M];
long long g[N + 1][M];
int main() {
  cin.tie(0), cout.tie(0);
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> l[i] >> r[i];
  for (int i = 1; i <= m; i++) {
    int res = m;
    vector<int> cnt;
    for (int j = 0; j < n; j++) {
      int dt = (l[j] + i - 1) / i;
      res -= dt * i;
      if (dt * i > r[j])
        res = -1;
      else
        cnt.push_back((r[j] - dt * i) / i + 1);
    }
    if (res < 0)
      f[i] = 0;
    else {
      int left = res / i;
      for (int j = 0; j <= left; j++) g[0][j] = 1;
      for (int k = 0; k < n; k++) {
        g[k + 1][0] = g[k][0];
        for (int j = 0; j <= left; j++) {
          g[k + 1][j] = g[k + 1][j - 1] + g[k][j];
          if (j - cnt[k] >= 0) g[k + 1][j] -= g[k][j - cnt[k]];
          g[k + 1][j] %= MOD;
        }
      }
      f[i] = g[n][left];
    }
  }
  for (int i = m; i >= 1; i--)
    for (int j = i + i; j <= m; j += i) f[i] = (f[i] - f[j]) % MOD;
  cout << (f[1] + MOD) % MOD << "\n";
}

### 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 int N = 50, M = 1e5 + 1, MOD = 998244353;
int l[N], r[N];
long long f[M];
long long g[N + 1][M];
int main() {
  cin.tie(0), cout.tie(0);
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> l[i] >> r[i];
  for (int i = 1; i <= m; i++) {
    int res = m;
    vector<int> cnt;
    for (int j = 0; j < n; j++) {
      int dt = (l[j] + i - 1) / i;
      res -= dt * i;
      if (dt * i > r[j])
        res = -1;
      else
        cnt.push_back((r[j] - dt * i) / i + 1);
    }
    if (res < 0)
      f[i] = 0;
    else {
      int left = res / i;
      for (int j = 0; j <= left; j++) g[0][j] = 1;
      for (int k = 0; k < n; k++) {
        g[k + 1][0] = g[k][0];
        for (int j = 0; j <= left; j++) {
          g[k + 1][j] = g[k + 1][j - 1] + g[k][j];
          if (j - cnt[k] >= 0) g[k + 1][j] -= g[k][j - cnt[k]];
          g[k + 1][j] %= MOD;
        }
      }
      f[i] = g[n][left];
    }
  }
  for (int i = m; i >= 1; i--)
    for (int j = i + i; j <= m; j += i) f[i] = (f[i] - f[j]) % MOD;
  cout << (f[1] + MOD) % MOD << "\n";
}

```