#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
  return x * f;
}
const long long mod = 998244353, N = 1e5 + 5;
long long p[N], chk[N], tot, mu[N];
void init(long long n) {
  mu[1] = 1;
  for (long long i = 2; i <= n; i++) {
    if (!chk[i]) p[++tot] = i, mu[i] = -1;
    for (long long j = 1; j <= tot && p[j] * i <= n; j++) {
      chk[i * p[j]] = 1;
      if (!(i % p[j])) break;
      mu[i * p[j]] = -mu[i];
    }
  }
}
long long ans;
long long l[N], r[N], f[N], s[N];
long long n, m;
signed main() {
  init(N - 5);
  n = read(), m = read();
  for (long long i = 1; i <= n; i++) l[i] = read(), r[i] = read();
  for (long long d = 1; d <= m; d++) {
    long long lim = m / d;
    for (long long i = 1; i <= lim; i++) f[i] = 0;
    f[0] = s[0] = 1;
    for (long long j = 1; j <= lim; j++) s[j] = (s[j - 1] + f[j]) % mod;
    for (long long i = 1; i <= n; i++) {
      long long L = (l[i] - 1) / d + 1, R = r[i] / d;
      for (long long j = 1; j <= lim; j++)
        f[j] = ((j - L < 0ll ? 0ll : s[max(j - L, 0ll)]) -
                (j - R - 1 < 0ll ? 0ll : s[max(j - R - 1, 0ll)])) %
               mod;
      f[0] = s[0] = 0;
      for (long long j = 1; j <= lim; j++) s[j] = (s[j - 1] + f[j]) % mod;
    }
    ans += mu[d] * s[lim];
    ans %= mod;
  }
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9')
    x = (x << 3) + (x << 1) + (c ^ '0'), c = getchar();
  return x * f;
}
const long long mod = 998244353, N = 1e5 + 5;
long long p[N], chk[N], tot, mu[N];
void init(long long n) {
  mu[1] = 1;
  for (long long i = 2; i <= n; i++) {
    if (!chk[i]) p[++tot] = i, mu[i] = -1;
    for (long long j = 1; j <= tot && p[j] * i <= n; j++) {
      chk[i * p[j]] = 1;
      if (!(i % p[j])) break;
      mu[i * p[j]] = -mu[i];
    }
  }
}
long long ans;
long long l[N], r[N], f[N], s[N];
long long n, m;
signed main() {
  init(N - 5);
  n = read(), m = read();
  for (long long i = 1; i <= n; i++) l[i] = read(), r[i] = read();
  for (long long d = 1; d <= m; d++) {
    long long lim = m / d;
    for (long long i = 1; i <= lim; i++) f[i] = 0;
    f[0] = s[0] = 1;
    for (long long j = 1; j <= lim; j++) s[j] = (s[j - 1] + f[j]) % mod;
    for (long long i = 1; i <= n; i++) {
      long long L = (l[i] - 1) / d + 1, R = r[i] / d;
      for (long long j = 1; j <= lim; j++)
        f[j] = ((j - L < 0ll ? 0ll : s[max(j - L, 0ll)]) -
                (j - R - 1 < 0ll ? 0ll : s[max(j - R - 1, 0ll)])) %
               mod;
      f[0] = s[0] = 0;
      for (long long j = 1; j <= lim; j++) s[j] = (s[j - 1] + f[j]) % mod;
    }
    ans += mu[d] * s[lim];
    ans %= mod;
  }
  printf("%lld\n", (ans + mod) % mod);
  return 0;
}

```