#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
using namespace std;
constexpr int N = 3e5 + 7;
constexpr int M = 998244353;
long long mu[N], flg[N], p[N], tot;
void getMu() {
  mu[1] = 1;
  for (long long i = 2; i < N; ++i) {
    if (!flg[i]) p[++tot] = i, mu[i] = -1;
    for (long long j = 1; j <= tot && i * p[j] < N; ++j) {
      flg[i * p[j]] = 1;
      if (i % p[j] == 0) {
        mu[i * p[j]] = 0;
        break;
      }
      mu[i * p[j]] = -mu[i];
    }
  }
}
long long dp[2][N];
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  getMu();
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, long long> > a(n + 1);
  for (long long i = 1; i <= n; i++) {
    cin >> a[i].first >> a[i].second;
  }
  long long ans = 0;
  for (long long d = 1; d <= m; d++) {
    fill(dp[0], dp[0] + m / d + 1, 1);
    long long x = 0;
    auto sum = [&](long long l, long long r) {
      return ((r >= 0 ? dp[x ^ 1][r] : 0ll) -
              (l - 1 >= 0 ? dp[x ^ 1][l - 1] : 0ll) + M) %
             M;
    };
    for (long long i = 1; i <= n; i++) {
      x ^= 1;
      fill(dp[x], dp[x] + m / d + 1, 0);
      long long l = a[i].first / d + (a[i].first % d != 0);
      long long r = a[i].second / d;
      for (long long j = 0; j <= m / d; j++) {
        dp[x][j] = sum(j - r, j - l);
        if (j - 1 >= 0) dp[x][j] = (dp[x][j] + dp[x][j - 1]) % M;
      }
    }
    ans = (ans + dp[x][m / d] * mu[d] % M + M) % M;
  }
  cout << ans << "\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>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC optimize("no-stack-protector,fast-math")
using namespace std;
constexpr int N = 3e5 + 7;
constexpr int M = 998244353;
long long mu[N], flg[N], p[N], tot;
void getMu() {
  mu[1] = 1;
  for (long long i = 2; i < N; ++i) {
    if (!flg[i]) p[++tot] = i, mu[i] = -1;
    for (long long j = 1; j <= tot && i * p[j] < N; ++j) {
      flg[i * p[j]] = 1;
      if (i % p[j] == 0) {
        mu[i * p[j]] = 0;
        break;
      }
      mu[i * p[j]] = -mu[i];
    }
  }
}
long long dp[2][N];
signed main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  getMu();
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, long long> > a(n + 1);
  for (long long i = 1; i <= n; i++) {
    cin >> a[i].first >> a[i].second;
  }
  long long ans = 0;
  for (long long d = 1; d <= m; d++) {
    fill(dp[0], dp[0] + m / d + 1, 1);
    long long x = 0;
    auto sum = [&](long long l, long long r) {
      return ((r >= 0 ? dp[x ^ 1][r] : 0ll) -
              (l - 1 >= 0 ? dp[x ^ 1][l - 1] : 0ll) + M) %
             M;
    };
    for (long long i = 1; i <= n; i++) {
      x ^= 1;
      fill(dp[x], dp[x] + m / d + 1, 0);
      long long l = a[i].first / d + (a[i].first % d != 0);
      long long r = a[i].second / d;
      for (long long j = 0; j <= m / d; j++) {
        dp[x][j] = sum(j - r, j - l);
        if (j - 1 >= 0) dp[x][j] = (dp[x][j] + dp[x][j - 1]) % M;
      }
    }
    ans = (ans + dp[x][m / d] * mu[d] % M + M) % M;
  }
  cout << ans << "\n";
  return 0;
}

```