#include <bits/stdc++.h>
const int maxn = 55, N = 1e5, mod = 998244353;
int n, m, l[maxn], r[maxn], vis[N + 5], pr[N + 5], mu[N + 5], prcnt;
void Sieve(int N) {
  mu[1] = 1;
  for (int i = 2; i <= N; ++i) {
    if (!vis[i]) pr[++prcnt] = i, mu[i] = -1;
    for (int j = 1; j <= prcnt && i * pr[j] <= N; ++j) {
      vis[i * pr[j]] = 1;
      if (i % pr[j] == 0) {
        mu[i * pr[j]] = 0;
        break;
      }
      mu[i * pr[j]] = -mu[i];
    }
  }
}
int f[N + 5], s[N + 5];
int Solve(int d) {
  f[0] = s[0] = 1;
  for (int i = 1; i <= m / d; ++i) f[i] = 0, s[i] = 1;
  for (int i = 1; i <= n; ++i) {
    int L = (l[i] - 1) / d + 1, R = r[i] / d;
    if (L > R) return 0;
    for (int j = 0; j < L; ++j) f[j] = 0;
    for (int j = L; j <= R; ++j) f[j] = s[j - L];
    for (int j = R + 1; j <= m / d; ++j)
      f[j] = (s[j - L] - s[j - R - 1] + mod) % mod;
    s[0] = f[0];
    for (int j = 1; j <= m / d; ++j) s[j] = (s[j - 1] + f[j]) % mod;
  }
  return s[m / d];
}
int main() {
  scanf("%d %d", &n, &m), Sieve(m);
  for (int i = 1; i <= n; ++i) scanf("%d %d", &l[i], &r[i]);
  int ans = 0;
  for (int i = 1; i <= m; ++i)
    mu[i] == 1 ? ans = (ans + Solve(i)) % mod
               : (mu[i] == -1 ? ans = (ans - Solve(i) + mod) % mod : 0);
  return printf("%d", ans), 0;
}

### Prompt

Create a solution in CPP for 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>
const int maxn = 55, N = 1e5, mod = 998244353;
int n, m, l[maxn], r[maxn], vis[N + 5], pr[N + 5], mu[N + 5], prcnt;
void Sieve(int N) {
  mu[1] = 1;
  for (int i = 2; i <= N; ++i) {
    if (!vis[i]) pr[++prcnt] = i, mu[i] = -1;
    for (int j = 1; j <= prcnt && i * pr[j] <= N; ++j) {
      vis[i * pr[j]] = 1;
      if (i % pr[j] == 0) {
        mu[i * pr[j]] = 0;
        break;
      }
      mu[i * pr[j]] = -mu[i];
    }
  }
}
int f[N + 5], s[N + 5];
int Solve(int d) {
  f[0] = s[0] = 1;
  for (int i = 1; i <= m / d; ++i) f[i] = 0, s[i] = 1;
  for (int i = 1; i <= n; ++i) {
    int L = (l[i] - 1) / d + 1, R = r[i] / d;
    if (L > R) return 0;
    for (int j = 0; j < L; ++j) f[j] = 0;
    for (int j = L; j <= R; ++j) f[j] = s[j - L];
    for (int j = R + 1; j <= m / d; ++j)
      f[j] = (s[j - L] - s[j - R - 1] + mod) % mod;
    s[0] = f[0];
    for (int j = 1; j <= m / d; ++j) s[j] = (s[j - 1] + f[j]) % mod;
  }
  return s[m / d];
}
int main() {
  scanf("%d %d", &n, &m), Sieve(m);
  for (int i = 1; i <= n; ++i) scanf("%d %d", &l[i], &r[i]);
  int ans = 0;
  for (int i = 1; i <= m; ++i)
    mu[i] == 1 ? ans = (ans + Solve(i)) % mod
               : (mu[i] == -1 ? ans = (ans - Solve(i) + mod) % mod : 0);
  return printf("%d", ans), 0;
}

```