#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int maxn = 1e5 + 5;
int cnt = 0, vis[maxn], prime[maxn], mu[maxn];
int f[55][maxn], l[55], r[55], l1[55], r1[55];
void init(int n) {
  vis[1] = mu[1] = 1;
  for (int i = (2); i <= (n); i++) {
    if (!vis[i]) prime[++cnt] = i, mu[i] = -1;
    for (int j = (1); j <= (cnt); j++) {
      if (i * prime[j] > n) break;
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}
int n, m;
int calc(int x) {
  long long sum = 0;
  for (int i = (1); i <= (n); i++) {
    l[i] = (l1[i] - 1) / x + 1;
    r[i] = r1[i] / x;
    sum += l[i];
    if (i == 1)
      for (int j = (l[i]); j <= (r[i]); j++) f[1][j] = 1;
    else {
      for (int j = (sum); j <= (m / x); j++) {
        int L = max(1, j - r[i]), R = j - l[i];
        f[i][j] = ((f[i - 1][R] - f[i - 1][L - 1]) % mod + mod) % mod;
      }
    }
    for (int j = (1); j <= (m / x); j++) (f[i][j] += f[i][j - 1]) %= mod;
  }
  return f[n][m / x];
}
int main() {
  init(1e5);
  scanf("%d", &n);
  scanf("%d", &m);
  int mnr = 1e9;
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &l1[i]), scanf("%d", &r1[i]);
    mnr = min(mnr, r1[i]);
  }
  long long ans = 0;
  for (int x = (1); x <= (mnr); x++) {
    ans = ((ans + calc(x) * mu[x]) % mod + mod) % mod;
  }
  cout << ans;
}

### 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 mod = 998244353;
const int maxn = 1e5 + 5;
int cnt = 0, vis[maxn], prime[maxn], mu[maxn];
int f[55][maxn], l[55], r[55], l1[55], r1[55];
void init(int n) {
  vis[1] = mu[1] = 1;
  for (int i = (2); i <= (n); i++) {
    if (!vis[i]) prime[++cnt] = i, mu[i] = -1;
    for (int j = (1); j <= (cnt); j++) {
      if (i * prime[j] > n) break;
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}
int n, m;
int calc(int x) {
  long long sum = 0;
  for (int i = (1); i <= (n); i++) {
    l[i] = (l1[i] - 1) / x + 1;
    r[i] = r1[i] / x;
    sum += l[i];
    if (i == 1)
      for (int j = (l[i]); j <= (r[i]); j++) f[1][j] = 1;
    else {
      for (int j = (sum); j <= (m / x); j++) {
        int L = max(1, j - r[i]), R = j - l[i];
        f[i][j] = ((f[i - 1][R] - f[i - 1][L - 1]) % mod + mod) % mod;
      }
    }
    for (int j = (1); j <= (m / x); j++) (f[i][j] += f[i][j - 1]) %= mod;
  }
  return f[n][m / x];
}
int main() {
  init(1e5);
  scanf("%d", &n);
  scanf("%d", &m);
  int mnr = 1e9;
  for (int i = (1); i <= (n); i++) {
    scanf("%d", &l1[i]), scanf("%d", &r1[i]);
    mnr = min(mnr, r1[i]);
  }
  long long ans = 0;
  for (int x = (1); x <= (mnr); x++) {
    ans = ((ans + calc(x) * mu[x]) % mod + mod) % mod;
  }
  cout << ans;
}

```