#include <bits/stdc++.h>
using namespace std;
using lli = long long int;
using ulli = unsigned long long int;
using Ld = long double;
using pii = pair<int, int>;
using pll = pair<lli, lli>;
using pld = pair<Ld, Ld>;
int pi[100001];
const lli MOD = 998244353;
pii p[51];
lli dp[51][100001];
lli GetRes(int n, int m, int g) {
  for (int i = (0); i <= (m); i++) dp[0][i] = 1;
  for (int k = (0); k < (n); k++) {
    lli l = (p[k].first + g - 1) / g, r = p[k].second / g;
    if (l > r) return 0;
    for (int i = (1); i <= (m); i++) {
      lli s = 0;
      if (i - l >= 0) s = dp[k][i - l];
      lli mi = 0;
      if (i - r - 1 >= 0) mi = dp[k][i - r - 1];
      dp[k + 1][i] = s - mi;
      dp[k + 1][i] = (dp[k + 1][i] + MOD) % MOD;
    }
    dp[k + 1][0] = 0;
    for (int i = (1); i <= (m); i++) {
      dp[k + 1][i] += dp[k + 1][i - 1];
      dp[k + 1][i] %= MOD;
    }
  }
  return dp[n][m];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = (0); i < (n); i++) cin >> p[i].first >> p[i].second;
  pi[1] = 1;
  lli res = 0;
  for (int i = (1); i <= (m / n + 1); i++) {
    if (pi[i] == 0) continue;
    res += pi[i] * GetRes(n, m / i, i);
    res = (res + MOD) % MOD;
    for (int j = i + i; j <= m / n + 1; j += i) pi[j] -= pi[i];
  }
  cout << res;
  cout.flush();
  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>
using namespace std;
using lli = long long int;
using ulli = unsigned long long int;
using Ld = long double;
using pii = pair<int, int>;
using pll = pair<lli, lli>;
using pld = pair<Ld, Ld>;
int pi[100001];
const lli MOD = 998244353;
pii p[51];
lli dp[51][100001];
lli GetRes(int n, int m, int g) {
  for (int i = (0); i <= (m); i++) dp[0][i] = 1;
  for (int k = (0); k < (n); k++) {
    lli l = (p[k].first + g - 1) / g, r = p[k].second / g;
    if (l > r) return 0;
    for (int i = (1); i <= (m); i++) {
      lli s = 0;
      if (i - l >= 0) s = dp[k][i - l];
      lli mi = 0;
      if (i - r - 1 >= 0) mi = dp[k][i - r - 1];
      dp[k + 1][i] = s - mi;
      dp[k + 1][i] = (dp[k + 1][i] + MOD) % MOD;
    }
    dp[k + 1][0] = 0;
    for (int i = (1); i <= (m); i++) {
      dp[k + 1][i] += dp[k + 1][i - 1];
      dp[k + 1][i] %= MOD;
    }
  }
  return dp[n][m];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = (0); i < (n); i++) cin >> p[i].first >> p[i].second;
  pi[1] = 1;
  lli res = 0;
  for (int i = (1); i <= (m / n + 1); i++) {
    if (pi[i] == 0) continue;
    res += pi[i] * GetRes(n, m / i, i);
    res = (res + MOD) % MOD;
    for (int j = i + i; j <= m / n + 1; j += i) pi[j] -= pi[i];
  }
  cout << res;
  cout.flush();
  return 0;
}

```