#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mod = 998244353;
inline int add(int x, int y) { return (x + y >= mod ? x + y - mod : x + y); }
inline int sub(int x, int y) { return (x - y < 0 ? x - y + mod : x - y); }
inline int gun(int x, int y) { return ((x * 1ll * y) % mod); }
int n, m;
pair<int, int> a[55];
int g[N];
int dp[N][55];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
  }
  for (int now = 1; now <= m; now++) {
    int sum = m / now;
    for (int idx = n; idx >= 0; idx--) {
      int l = (a[idx].first + now - 1) / now;
      int r = a[idx].second / now;
      for (int cur = 0; cur <= sum; cur++) {
        if (idx == n) {
          dp[cur][idx] = 1;
          continue;
        }
        int ans = 0;
        if (l <= r && cur + l <= sum) {
          int ses = min(cur + r, sum);
          ans = dp[ses][idx + 1];
          if (cur + l - 1 >= 0) ans = sub(ans, dp[cur + l - 1][idx + 1]);
        }
        dp[cur][idx] = ans;
      }
      for (int cur = 1; cur <= sum; cur++) {
        dp[cur][idx] = add(dp[cur][idx], dp[cur - 1][idx]);
      }
    }
    g[now] = dp[0][0];
    ;
  }
  for (int i = m; i >= 1; i--) {
    for (int j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        if (j != i) g[j] = sub(g[j], g[i]);
        int arek = i / j;
        if (arek != i && arek != j) g[arek] = sub(g[arek], g[i]);
      }
    }
  }
  cout << g[1] << "\n";
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 N = 1e5 + 5;
const int mod = 998244353;
inline int add(int x, int y) { return (x + y >= mod ? x + y - mod : x + y); }
inline int sub(int x, int y) { return (x - y < 0 ? x - y + mod : x - y); }
inline int gun(int x, int y) { return ((x * 1ll * y) % mod); }
int n, m;
pair<int, int> a[55];
int g[N];
int dp[N][55];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i].first >> a[i].second;
  }
  for (int now = 1; now <= m; now++) {
    int sum = m / now;
    for (int idx = n; idx >= 0; idx--) {
      int l = (a[idx].first + now - 1) / now;
      int r = a[idx].second / now;
      for (int cur = 0; cur <= sum; cur++) {
        if (idx == n) {
          dp[cur][idx] = 1;
          continue;
        }
        int ans = 0;
        if (l <= r && cur + l <= sum) {
          int ses = min(cur + r, sum);
          ans = dp[ses][idx + 1];
          if (cur + l - 1 >= 0) ans = sub(ans, dp[cur + l - 1][idx + 1]);
        }
        dp[cur][idx] = ans;
      }
      for (int cur = 1; cur <= sum; cur++) {
        dp[cur][idx] = add(dp[cur][idx], dp[cur - 1][idx]);
      }
    }
    g[now] = dp[0][0];
    ;
  }
  for (int i = m; i >= 1; i--) {
    for (int j = 1; j * j <= i; j++) {
      if (i % j == 0) {
        if (j != i) g[j] = sub(g[j], g[i]);
        int arek = i / j;
        if (arek != i && arek != j) g[arek] = sub(g[arek], g[i]);
      }
    }
  }
  cout << g[1] << "\n";
  return 0;
}

```