#include <bits/stdc++.h>
using namespace std;
const int N = 57, M = 1e5 + 7, MOD = 998244353;
int f[N][M], s[M];
int _L[N], _R[N], L[N], R[N];
int primes[M], cnt, mob[M];
bool st[M];
void init() {
  mob[1] = 1;
  for (int i = 2; i <= M - 1; ++i) {
    if (!st[i]) primes[cnt++] = i, mob[i] = -1;
    for (int j = 0; j < cnt && 1ll * primes[j] * i < M; ++j) {
      st[primes[j] * i] = 1;
      if (i % primes[j] == 0) break;
      mob[i * primes[j]] = -mob[i];
    }
  }
}
int dp(int n, int m) {
  f[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= m; ++j)
      s[j] = ((j > 0 ? s[j - 1] : 0) + f[i - 1][j]) % MOD;
    for (int j = L[i]; j <= m; ++j)
      f[i][j] = (s[j - L[i]] - (j - R[i] <= 0 ? 0 : s[j - R[i] - 1])) % MOD;
  }
  int res = 0;
  for (int j = 0; j <= m; ++j) res = (res + f[n][j]) % MOD;
  return res;
}
int main() {
  init();
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &_L[i], &_R[i]);
  int res = 0;
  for (int d = 1; d <= m; ++d) {
    for (int i = 1; i <= n; ++i) L[i] = (_L[i] + d - 1) / d, R[i] = _R[i] / d;
    res = (res + 1ll * mob[d] * dp(n, m / d) % MOD) % MOD;
    res = (res % MOD + MOD) % MOD;
  }
  printf("%d\n", res);
  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 = 57, M = 1e5 + 7, MOD = 998244353;
int f[N][M], s[M];
int _L[N], _R[N], L[N], R[N];
int primes[M], cnt, mob[M];
bool st[M];
void init() {
  mob[1] = 1;
  for (int i = 2; i <= M - 1; ++i) {
    if (!st[i]) primes[cnt++] = i, mob[i] = -1;
    for (int j = 0; j < cnt && 1ll * primes[j] * i < M; ++j) {
      st[primes[j] * i] = 1;
      if (i % primes[j] == 0) break;
      mob[i * primes[j]] = -mob[i];
    }
  }
}
int dp(int n, int m) {
  f[0][0] = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= m; ++j)
      s[j] = ((j > 0 ? s[j - 1] : 0) + f[i - 1][j]) % MOD;
    for (int j = L[i]; j <= m; ++j)
      f[i][j] = (s[j - L[i]] - (j - R[i] <= 0 ? 0 : s[j - R[i] - 1])) % MOD;
  }
  int res = 0;
  for (int j = 0; j <= m; ++j) res = (res + f[n][j]) % MOD;
  return res;
}
int main() {
  init();
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &_L[i], &_R[i]);
  int res = 0;
  for (int d = 1; d <= m; ++d) {
    for (int i = 1; i <= n; ++i) L[i] = (_L[i] + d - 1) / d, R[i] = _R[i] / d;
    res = (res + 1ll * mob[d] * dp(n, m / d) % MOD) % MOD;
    res = (res % MOD + MOD) % MOD;
  }
  printf("%d\n", res);
  return 0;
}

```