#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const int N = 1e5;
const int M = 50;
int num[N + 5];
int mob[N + 5];
void pre() {
  mob[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (num[i] == 0) {
      for (int j = i; j <= N; j += i) {
        mob[j]++;
        num[j] = j;
      }
      long long x = i;
      x = x * i;
      for (long long j = x; j <= N; j += x) mob[j] = -N;
    }
    if (mob[i] < 0)
      mob[i] = 0;
    else if (mob[i] % 2)
      mob[i] = -1;
    else
      mob[i] = 1;
  }
}
long long dp[M + 1][N + 1];
long long ssum[N + 5];
int l[M + 1], r[M + 1];
int nr[M + 1];
inline long long get(int l, int r) {
  if (l < 0) l = 0;
  return (ssum[l] - ssum[r + 1] + MOD) % MOD;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  pre();
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
  long long res = 0;
  for (int i = 1; i <= m; i++) {
    int sisa = m / i;
    bool bisa = true;
    for (int j = 1; j <= n; j++) {
      int nl = l[j] / i;
      if (nl * i < l[j]) nl++;
      nr[j] = r[j] / i;
      if (nr[j] * i < l[j] || nl * i > r[j] || nl > nr[j]) bisa = false;
      sisa -= nl;
      nr[j] -= nl;
    }
    if (mob[i] == 0 || !bisa || sisa < 0) continue;
    for (int j = 0; j <= max(sisa, nr[j]) + 1; j++) ssum[j] = 0;
    for (int j = min(sisa, nr[1]); j >= 0; j--) {
      dp[1][j] = 1;
      ssum[j] = (ssum[j + 1] + dp[1][j]) % MOD;
    }
    for (int j = 2; j <= n; j++) {
      for (int k = sisa; k >= 0; k--) {
        dp[j][k] = get(k - nr[j], k);
      }
      for (int k = sisa; k >= 0; k--) {
        ssum[k] = (ssum[k + 1] + dp[j][k]) % MOD;
      }
    }
    res = (res + MOD + mob[i] * ssum[0]) % MOD;
  }
  cout << res << '\n';
}

### Prompt

Your challenge is to write 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 long long MOD = 998244353;
const int N = 1e5;
const int M = 50;
int num[N + 5];
int mob[N + 5];
void pre() {
  mob[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (num[i] == 0) {
      for (int j = i; j <= N; j += i) {
        mob[j]++;
        num[j] = j;
      }
      long long x = i;
      x = x * i;
      for (long long j = x; j <= N; j += x) mob[j] = -N;
    }
    if (mob[i] < 0)
      mob[i] = 0;
    else if (mob[i] % 2)
      mob[i] = -1;
    else
      mob[i] = 1;
  }
}
long long dp[M + 1][N + 1];
long long ssum[N + 5];
int l[M + 1], r[M + 1];
int nr[M + 1];
inline long long get(int l, int r) {
  if (l < 0) l = 0;
  return (ssum[l] - ssum[r + 1] + MOD) % MOD;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  pre();
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
  long long res = 0;
  for (int i = 1; i <= m; i++) {
    int sisa = m / i;
    bool bisa = true;
    for (int j = 1; j <= n; j++) {
      int nl = l[j] / i;
      if (nl * i < l[j]) nl++;
      nr[j] = r[j] / i;
      if (nr[j] * i < l[j] || nl * i > r[j] || nl > nr[j]) bisa = false;
      sisa -= nl;
      nr[j] -= nl;
    }
    if (mob[i] == 0 || !bisa || sisa < 0) continue;
    for (int j = 0; j <= max(sisa, nr[j]) + 1; j++) ssum[j] = 0;
    for (int j = min(sisa, nr[1]); j >= 0; j--) {
      dp[1][j] = 1;
      ssum[j] = (ssum[j + 1] + dp[1][j]) % MOD;
    }
    for (int j = 2; j <= n; j++) {
      for (int k = sisa; k >= 0; k--) {
        dp[j][k] = get(k - nr[j], k);
      }
      for (int k = sisa; k >= 0; k--) {
        ssum[k] = (ssum[k + 1] + dp[j][k]) % MOD;
      }
    }
    res = (res + MOD + mob[i] * ssum[0]) % MOD;
  }
  cout << res << '\n';
}

```