#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
const char nl = '\n';
const ll mod = 998244353;
const int max_n = 50;
const int max_m = 1e5;
const int max_sum = max_n * max_m;
int l[max_n + 1], r[max_n + 1];
ll f[max_m + 1];
ll cnt[max_m + 2][max_n + 1];
ll cnt_pref[max_m + 2][max_n + 1];
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
  int l_sum = 0;
  for (int i = 1; i <= n; i++) l_sum += l[i];
  if (l_sum > m) {
    cout << 0 << nl;
    return;
  }
  for (int d = m / n; d >= 1; d--) {
    int ms = m / d;
    bool can = true;
    for (int i = 1; i <= n; i++) {
      can &= (l[i] + d - 1) / d * d <= r[i];
    }
    if (!can) continue;
    for (int i = 0; i <= n; i++) {
      for (int s = 0; s <= ms; s++) {
        cnt[s][i] = cnt_pref[s][i] = 0;
      }
    }
    cnt[0][0] = 1;
    for (int s = 0; s <= ms; s++) cnt_pref[s][0] = 1;
    for (int i = 1; i <= n; i++) {
      for (int s = 0; s <= ms; s++) {
        int left = max(0, s - r[i] / d), right = s - (l[i] + d - 1) / d;
        if (left > right) continue;
        ll range_sum = cnt_pref[right][i - 1];
        if (left > 0) {
          range_sum -= cnt_pref[left - 1][i - 1];
          range_sum += mod;
          range_sum %= mod;
        }
        cnt[s][i] = range_sum;
        cnt_pref[s][i] = cnt[s][i];
        if (s > 0) {
          cnt_pref[s][i] += cnt_pref[s - 1][i];
          cnt_pref[s][i] %= mod;
        }
      }
    }
    f[d] = cnt_pref[ms][n];
    for (int d2 = 2 * d; d2 <= m; d2 += d) {
      f[d] -= f[d2];
      f[d] += mod;
      f[d] %= mod;
    }
  }
  cout << f[1] << nl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) solve();
  return 0;
}

### 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;
using ll = long long;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vl = vector<ll>;
const char nl = '\n';
const ll mod = 998244353;
const int max_n = 50;
const int max_m = 1e5;
const int max_sum = max_n * max_m;
int l[max_n + 1], r[max_n + 1];
ll f[max_m + 1];
ll cnt[max_m + 2][max_n + 1];
ll cnt_pref[max_m + 2][max_n + 1];
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
  int l_sum = 0;
  for (int i = 1; i <= n; i++) l_sum += l[i];
  if (l_sum > m) {
    cout << 0 << nl;
    return;
  }
  for (int d = m / n; d >= 1; d--) {
    int ms = m / d;
    bool can = true;
    for (int i = 1; i <= n; i++) {
      can &= (l[i] + d - 1) / d * d <= r[i];
    }
    if (!can) continue;
    for (int i = 0; i <= n; i++) {
      for (int s = 0; s <= ms; s++) {
        cnt[s][i] = cnt_pref[s][i] = 0;
      }
    }
    cnt[0][0] = 1;
    for (int s = 0; s <= ms; s++) cnt_pref[s][0] = 1;
    for (int i = 1; i <= n; i++) {
      for (int s = 0; s <= ms; s++) {
        int left = max(0, s - r[i] / d), right = s - (l[i] + d - 1) / d;
        if (left > right) continue;
        ll range_sum = cnt_pref[right][i - 1];
        if (left > 0) {
          range_sum -= cnt_pref[left - 1][i - 1];
          range_sum += mod;
          range_sum %= mod;
        }
        cnt[s][i] = range_sum;
        cnt_pref[s][i] = cnt[s][i];
        if (s > 0) {
          cnt_pref[s][i] += cnt_pref[s - 1][i];
          cnt_pref[s][i] %= mod;
        }
      }
    }
    f[d] = cnt_pref[ms][n];
    for (int d2 = 2 * d; d2 <= m; d2 += d) {
      f[d] -= f[d2];
      f[d] += mod;
      f[d] %= mod;
    }
  }
  cout << f[1] << nl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t = 1;
  while (t--) solve();
  return 0;
}

```