#include <bits/stdc++.h>
using namespace std;
const int N = 50 + 5;
const long long M = 1e5 + 5;
const long long mod = 998244353;
const int inf = 1e9;
const double eps = 1e-9;
const double PI = acos(-1.0);
const pair<int, int> NIL = {0, 0};
int n, m;
long long l[N], r[N], mu[M], sum[M], f[M];
long long deal(long long d) {
  long long lim = m / d;
  f[0] = 1;
  for (long long i = 1; i <= lim; ++i) f[i] = 0;
  for (long long i = 1; i <= n; ++i) {
    long long L = (l[i] + d - 1) / d, R = r[i] / d;
    if (L > R) return 0;
    sum[0] = f[0];
    for (long long j = 1; j <= lim; ++j) sum[j] = (f[j] + sum[j - 1]) % mod;
    for (long long j = 0; j <= lim; ++j) {
      if (j > R)
        f[j] = (sum[j - L] - sum[j - R - 1] + mod) % mod;
      else if (j >= L)
        f[j] = sum[j - L];
      else
        f[j] = 0;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= lim; ++i) ans = (ans + f[i]) % mod;
  return ans;
}
void solve() {
  mu[1] = 1;
  cin >> n >> m;
  for (long long i = 1; i <= m; ++i)
    for (int j = i + i; j <= m; j += i) mu[j] = (mu[j] + mod - mu[i]) % mod;
  for (long long i = 1; i <= n; ++i) cin >> l[i] >> r[i];
  long long ans = 0;
  for (long long i = 1; i <= m; ++i) {
    ans = (mu[i] * deal(i) + ans + mod) % mod;
  }
  cout << ans << "\n";
}
int main() {
  solve();
  return 0;
}

### Prompt

Please formulate 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 int N = 50 + 5;
const long long M = 1e5 + 5;
const long long mod = 998244353;
const int inf = 1e9;
const double eps = 1e-9;
const double PI = acos(-1.0);
const pair<int, int> NIL = {0, 0};
int n, m;
long long l[N], r[N], mu[M], sum[M], f[M];
long long deal(long long d) {
  long long lim = m / d;
  f[0] = 1;
  for (long long i = 1; i <= lim; ++i) f[i] = 0;
  for (long long i = 1; i <= n; ++i) {
    long long L = (l[i] + d - 1) / d, R = r[i] / d;
    if (L > R) return 0;
    sum[0] = f[0];
    for (long long j = 1; j <= lim; ++j) sum[j] = (f[j] + sum[j - 1]) % mod;
    for (long long j = 0; j <= lim; ++j) {
      if (j > R)
        f[j] = (sum[j - L] - sum[j - R - 1] + mod) % mod;
      else if (j >= L)
        f[j] = sum[j - L];
      else
        f[j] = 0;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= lim; ++i) ans = (ans + f[i]) % mod;
  return ans;
}
void solve() {
  mu[1] = 1;
  cin >> n >> m;
  for (long long i = 1; i <= m; ++i)
    for (int j = i + i; j <= m; j += i) mu[j] = (mu[j] + mod - mu[i]) % mod;
  for (long long i = 1; i <= n; ++i) cin >> l[i] >> r[i];
  long long ans = 0;
  for (long long i = 1; i <= m; ++i) {
    ans = (mu[i] * deal(i) + ans + mod) % mod;
  }
  cout << ans << "\n";
}
int main() {
  solve();
  return 0;
}

```