#include <bits/stdc++.h>
using namespace std;
const long long MOD = 998244353;
const int N = 20000;
const int LG = 32 - __builtin_clz(N);
const int INF = 1e9;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > ran(n);
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    ran[i] = {l, r};
  }
  int gcd = m / n;
  vector<long long> ans(gcd + 1, 0);
  for (int div = 1; div <= gcd; div++) {
    int nm = m / div;
    vector<long long> dp(nm + 1, 1);
    vector<long long> sp(nm + 2, 0);
    for (int i = n - 1; i >= 0; i--) {
      for (int j = 0; j <= nm; j++) {
        sp[j + 1] = (dp[j] + sp[j]) % MOD;
      }
      int l = (ran[i].first + div - 1) / div;
      int r = ran[i].second / div;
      if (l > r) continue;
      for (int amo = 0; amo < l; amo++) dp[amo] = 0;
      for (int amo = l; amo <= nm; amo++) {
        int top = amo - l;
        int bot = max(0, amo - r);
        dp[amo] = (sp[top + 1] - sp[bot] + MOD) % MOD;
      }
    }
    ans[div] = dp[nm];
  }
  for (int i = gcd; i > 0; i--) {
    for (int j = 2 * i; j <= gcd; j += i) {
      ans[i] = (ans[i] - ans[j] + MOD) % MOD;
    }
  }
  if (gcd == 0)
    cout << 0 << '\n';
  else
    cout << ans[1] << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int tt = 1;
  while (tt--) solve();
  return 0;
}

### Prompt

Generate 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 = 20000;
const int LG = 32 - __builtin_clz(N);
const int INF = 1e9;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > ran(n);
  for (int i = 0; i < n; i++) {
    int l, r;
    cin >> l >> r;
    ran[i] = {l, r};
  }
  int gcd = m / n;
  vector<long long> ans(gcd + 1, 0);
  for (int div = 1; div <= gcd; div++) {
    int nm = m / div;
    vector<long long> dp(nm + 1, 1);
    vector<long long> sp(nm + 2, 0);
    for (int i = n - 1; i >= 0; i--) {
      for (int j = 0; j <= nm; j++) {
        sp[j + 1] = (dp[j] + sp[j]) % MOD;
      }
      int l = (ran[i].first + div - 1) / div;
      int r = ran[i].second / div;
      if (l > r) continue;
      for (int amo = 0; amo < l; amo++) dp[amo] = 0;
      for (int amo = l; amo <= nm; amo++) {
        int top = amo - l;
        int bot = max(0, amo - r);
        dp[amo] = (sp[top + 1] - sp[bot] + MOD) % MOD;
      }
    }
    ans[div] = dp[nm];
  }
  for (int i = gcd; i > 0; i--) {
    for (int j = 2 * i; j <= gcd; j += i) {
      ans[i] = (ans[i] - ans[j] + MOD) % MOD;
    }
  }
  if (gcd == 0)
    cout << 0 << '\n';
  else
    cout << ans[1] << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int tt = 1;
  while (tt--) solve();
  return 0;
}

```