#include <bits/stdc++.h>
using namespace std;
const int N = 50 + 13, M = 1e5 + 13, mod = 998244353;
int n, m, prm[M], cnt, mu[M], L[N], R[N], l[N], r[N], f[M], g[M];
bool b[M];
inline void init() {
  mu[1] = 1;
  for (int i = 2; i <= m; ++i) {
    if (!b[i]) prm[++cnt] = i, mu[i] = -1;
    for (int j = 1; j <= cnt && i * prm[j] <= m; ++j) {
      b[i * prm[j]] = 1;
      if (i % prm[j] == 0) break;
      mu[i * prm[j]] = -mu[i];
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &L[i], &R[i]);
  init();
  int ans = 0;
  for (int d = 1; d <= m; ++d) {
    if (!mu[d]) continue;
    for (int i = 1; i <= n; ++i) l[i] = (L[i] - 1) / d + 1, r[i] = R[i] / d;
    int tmp = m / d;
    for (int i = 1; i <= tmp; ++i) f[i] = 0;
    for (int i = l[1]; i <= r[1]; ++i) f[i] = 1;
    for (int i = 1; i <= tmp; ++i) g[i] = (g[i - 1] + f[i]) % mod;
    for (int i = 2; i <= n; ++i) {
      for (int j = 1; j < l[i]; ++j) f[j] = 0;
      for (int j = l[i]; j <= tmp; ++j) {
        f[j] = g[j - l[i]];
        if (j - r[i] - 1 >= 1)
          f[j] -= g[j - r[i] - 1], f[j] = (f[j] % mod + mod) % mod;
      }
      g[0] = 0;
      for (int j = 1; j <= tmp; ++j) g[j] = (g[j - 1] + f[j]) % mod;
    }
    ans += mu[d] * g[tmp] % mod, ans = (ans % mod + mod) % mod;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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 + 13, M = 1e5 + 13, mod = 998244353;
int n, m, prm[M], cnt, mu[M], L[N], R[N], l[N], r[N], f[M], g[M];
bool b[M];
inline void init() {
  mu[1] = 1;
  for (int i = 2; i <= m; ++i) {
    if (!b[i]) prm[++cnt] = i, mu[i] = -1;
    for (int j = 1; j <= cnt && i * prm[j] <= m; ++j) {
      b[i * prm[j]] = 1;
      if (i % prm[j] == 0) break;
      mu[i * prm[j]] = -mu[i];
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &L[i], &R[i]);
  init();
  int ans = 0;
  for (int d = 1; d <= m; ++d) {
    if (!mu[d]) continue;
    for (int i = 1; i <= n; ++i) l[i] = (L[i] - 1) / d + 1, r[i] = R[i] / d;
    int tmp = m / d;
    for (int i = 1; i <= tmp; ++i) f[i] = 0;
    for (int i = l[1]; i <= r[1]; ++i) f[i] = 1;
    for (int i = 1; i <= tmp; ++i) g[i] = (g[i - 1] + f[i]) % mod;
    for (int i = 2; i <= n; ++i) {
      for (int j = 1; j < l[i]; ++j) f[j] = 0;
      for (int j = l[i]; j <= tmp; ++j) {
        f[j] = g[j - l[i]];
        if (j - r[i] - 1 >= 1)
          f[j] -= g[j - r[i] - 1], f[j] = (f[j] % mod + mod) % mod;
      }
      g[0] = 0;
      for (int j = 1; j <= tmp; ++j) g[j] = (g[j - 1] + f[j]) % mod;
    }
    ans += mu[d] * g[tmp] % mod, ans = (ans % mod + mod) % mod;
  }
  printf("%d\n", ans);
  return 0;
}

```