#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
int p[2][100001];
int get(int i, int l, int r) {
  if (r < 0) return 0;
  l = max(l, 0);
  return add(p[i][r], mod - (l == 0 ? 0 : p[i][l - 1]));
}
void solution() {
  int n, m;
  cin >> n >> m;
  vector<int> f(m + 1);
  vector<int> l(n), r(n);
  for (int i = 0; i < n; i++) cin >> l[i] >> r[i];
  for (int d = 1; d <= m; d++) {
    vector<int> L(n), R(n);
    bool fail = false;
    for (int i = 0; i < n; i++) {
      L[i] = (l[i] + d - 1) / d;
      R[i] = r[i] / d;
      if (R[i] < L[i]) fail = true;
    }
    if (fail) continue;
    int bound = m / d;
    for (int i = 0; i <= 1; i++)
      for (int j = 0; j <= bound; j++) p[i][j] = 0;
    p[0][0] = 1;
    for (int i = 1; i <= bound; i++) p[0][i] = 1;
    int cur = 0, nxt = 1;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= bound; j++)
        p[nxt][j] = add(p[nxt][j], get(cur, j - R[i], j - L[i]));
      for (int j = 1; j <= bound; j++)
        p[nxt][j] = add(p[nxt][j], p[nxt][j - 1]);
      for (int j = 0; j <= bound; j++) p[cur][j] = 0;
      swap(cur, nxt);
    }
    f[d] = p[cur][bound];
  }
  for (int i = m; i >= 1; i--) {
    for (int j = i + i; j <= m; j += i) f[i] = add(f[i], mod - f[j]);
  }
  cout << f[1];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) {
    solution();
    cout << "\n";
  }
}

### 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 mod = 998244353;
int add(int x, int y) {
  x += y;
  if (x >= mod) x -= mod;
  return x;
}
int p[2][100001];
int get(int i, int l, int r) {
  if (r < 0) return 0;
  l = max(l, 0);
  return add(p[i][r], mod - (l == 0 ? 0 : p[i][l - 1]));
}
void solution() {
  int n, m;
  cin >> n >> m;
  vector<int> f(m + 1);
  vector<int> l(n), r(n);
  for (int i = 0; i < n; i++) cin >> l[i] >> r[i];
  for (int d = 1; d <= m; d++) {
    vector<int> L(n), R(n);
    bool fail = false;
    for (int i = 0; i < n; i++) {
      L[i] = (l[i] + d - 1) / d;
      R[i] = r[i] / d;
      if (R[i] < L[i]) fail = true;
    }
    if (fail) continue;
    int bound = m / d;
    for (int i = 0; i <= 1; i++)
      for (int j = 0; j <= bound; j++) p[i][j] = 0;
    p[0][0] = 1;
    for (int i = 1; i <= bound; i++) p[0][i] = 1;
    int cur = 0, nxt = 1;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= bound; j++)
        p[nxt][j] = add(p[nxt][j], get(cur, j - R[i], j - L[i]));
      for (int j = 1; j <= bound; j++)
        p[nxt][j] = add(p[nxt][j], p[nxt][j - 1]);
      for (int j = 0; j <= bound; j++) p[cur][j] = 0;
      swap(cur, nxt);
    }
    f[d] = p[cur][bound];
  }
  for (int i = m; i >= 1; i--) {
    for (int j = i + i; j <= m; j += i) f[i] = add(f[i], mod - f[j]);
  }
  cout << f[1];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  while (tc--) {
    solution();
    cout << "\n";
  }
}

```