#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const int N = 50;
const int MX = 1e5;
const int MOD = 998244353;
int n, m;
int L[N + 5], R[N + 5];
int up[N + 5], low[N + 5];
int f[MX + 5];
inline void add(int& a, int b) { a = (a + b) % MOD; }
inline int sub(int a, int b) {
  a = (a - b) % MOD;
  if (a < 0) a += MOD;
  return a;
}
int calc(int k) {
  vector<vector<int>> dp(n, vector<int>(m / k + 5));
  vector<int> pre(m / k + 5);
  int bat = m / k;
  for (int i = 0; i < n; i++) {
    low[i] = ((L[i] + k - 1) / k);
    bat -= (low[i]);
    up[i] = ((R[i]) / k);
  }
  if (bat < 0) return 0;
  for (int i = 0; i <= up[0] - low[0]; i++) {
    dp[0][i] = 1;
  }
  pre[0] = 1;
  for (int i = 1; i <= bat; i++) {
    pre[i] = (pre[i - 1] + dp[0][i]) % MOD;
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= bat; j++) {
      int lul;
      if (j > (up[i] - low[i]))
        lul = pre[j - (up[i] - low[i]) - 1];
      else
        lul = 0;
      dp[i][j] = (dp[i][j] + sub(pre[j], lul)) % MOD;
    }
    pre[0] = dp[i][0];
    for (int j = 1; j <= bat; j++) {
      pre[j] = (pre[j - 1] + dp[i][j]) % MOD;
    }
  }
  int ret = 0;
  for (int i = 0; i <= bat; i++) {
    ret = (ret + dp[n - 1][i]) % MOD;
  }
  return ret;
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> L[i] >> R[i];
  }
  for (int i = 1; i <= m / n; i++) {
    f[i] = calc(i);
  }
  for (int i = m; i >= 1; i--) {
    for (int j = i + i; j <= m; j += i) {
      f[i] = (f[i] - f[j] + MOD) % MOD;
    }
  }
  int ans = f[1] % MOD;
  if (ans < 0) ans += MOD;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TC = 1;
  for (int i = 1; i <= TC; i++) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
using ll = long long;
using ld = long double;
const int N = 50;
const int MX = 1e5;
const int MOD = 998244353;
int n, m;
int L[N + 5], R[N + 5];
int up[N + 5], low[N + 5];
int f[MX + 5];
inline void add(int& a, int b) { a = (a + b) % MOD; }
inline int sub(int a, int b) {
  a = (a - b) % MOD;
  if (a < 0) a += MOD;
  return a;
}
int calc(int k) {
  vector<vector<int>> dp(n, vector<int>(m / k + 5));
  vector<int> pre(m / k + 5);
  int bat = m / k;
  for (int i = 0; i < n; i++) {
    low[i] = ((L[i] + k - 1) / k);
    bat -= (low[i]);
    up[i] = ((R[i]) / k);
  }
  if (bat < 0) return 0;
  for (int i = 0; i <= up[0] - low[0]; i++) {
    dp[0][i] = 1;
  }
  pre[0] = 1;
  for (int i = 1; i <= bat; i++) {
    pre[i] = (pre[i - 1] + dp[0][i]) % MOD;
  }
  for (int i = 1; i < n; i++) {
    for (int j = 0; j <= bat; j++) {
      int lul;
      if (j > (up[i] - low[i]))
        lul = pre[j - (up[i] - low[i]) - 1];
      else
        lul = 0;
      dp[i][j] = (dp[i][j] + sub(pre[j], lul)) % MOD;
    }
    pre[0] = dp[i][0];
    for (int j = 1; j <= bat; j++) {
      pre[j] = (pre[j - 1] + dp[i][j]) % MOD;
    }
  }
  int ret = 0;
  for (int i = 0; i <= bat; i++) {
    ret = (ret + dp[n - 1][i]) % MOD;
  }
  return ret;
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> L[i] >> R[i];
  }
  for (int i = 1; i <= m / n; i++) {
    f[i] = calc(i);
  }
  for (int i = m; i >= 1; i--) {
    for (int j = i + i; j <= m; j += i) {
      f[i] = (f[i] - f[j] + MOD) % MOD;
    }
  }
  int ans = f[1] % MOD;
  if (ans < 0) ans += MOD;
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int TC = 1;
  for (int i = 1; i <= TC; i++) {
    solve();
  }
  return 0;
}

```