#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 998244353;
double eps = 1e-12;
const int N = 3e5 + 7;
int n, m;
int mnrr[57], mxrr[57];
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> mnrr[i] >> mxrr[i];
  vector<long long> ans(m + 1, 0);
  for (int g = 1; g <= m; g++) {
    long long limit = m / g;
    vector<long long> dp(limit + 1, 1);
    for (int i = 1; i <= n; i++) {
      long long rngmn = (mnrr[i] + g - 1) / g;
      long long rngmx = mxrr[i] / g;
      if (rngmn > rngmx) {
        dp[limit] = 0;
        break;
      }
      vector<long long> ndp(limit + 1, 0);
      for (int sum = rngmn; sum <= limit; sum++) {
        ndp[sum] = dp[sum - rngmn];
        if (sum > rngmx) {
          ndp[sum] = (ndp[sum] - dp[sum - rngmx - 1]) % MOD;
        }
        ndp[sum] = (ndp[sum - 1] + ndp[sum]) % MOD;
      }
      dp = ndp;
    }
    ans[g] = dp[limit];
  }
  for (int i = m; i >= 1; i--) {
    for (int j = i + i; j <= m; j += i) {
      ans[i] = (ans[i] - ans[j] + MOD) % MOD;
    }
  }
  cout << ans[1] << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int it = 1; it <= t; it++) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 998244353;
double eps = 1e-12;
const int N = 3e5 + 7;
int n, m;
int mnrr[57], mxrr[57];
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> mnrr[i] >> mxrr[i];
  vector<long long> ans(m + 1, 0);
  for (int g = 1; g <= m; g++) {
    long long limit = m / g;
    vector<long long> dp(limit + 1, 1);
    for (int i = 1; i <= n; i++) {
      long long rngmn = (mnrr[i] + g - 1) / g;
      long long rngmx = mxrr[i] / g;
      if (rngmn > rngmx) {
        dp[limit] = 0;
        break;
      }
      vector<long long> ndp(limit + 1, 0);
      for (int sum = rngmn; sum <= limit; sum++) {
        ndp[sum] = dp[sum - rngmn];
        if (sum > rngmx) {
          ndp[sum] = (ndp[sum] - dp[sum - rngmx - 1]) % MOD;
        }
        ndp[sum] = (ndp[sum - 1] + ndp[sum]) % MOD;
      }
      dp = ndp;
    }
    ans[g] = dp[limit];
  }
  for (int i = m; i >= 1; i--) {
    for (int j = i + i; j <= m; j += i) {
      ans[i] = (ans[i] - ans[j] + MOD) % MOD;
    }
  }
  cout << ans[1] << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int it = 1; it <= t; it++) {
    solve();
  }
  return 0;
}

```