#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 10, mod = 998244353;
long long prim[N], nump, mu[N], vis[N], l[N], r[N];
long long f[55][N], sum[N];
int n, m;
void init() {
  mu[1] = 1;
  for (int i = 2; i <= N - 1; i++) {
    if (!vis[i]) {
      prim[++nump] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= nump && i * prim[j] <= N - 1; j++) {
      vis[i * prim[j]] = 1;
      if (i % prim[j] == 0)
        break;
      else
        mu[i * prim[j]] = -mu[i];
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  init();
  for (int i = 1; i <= n; i++) scanf("%d%d", l + i, r + i);
  long long ans = 0;
  for (int d = 1; d <= m; d++) {
    f[0][0] = 1;
    long long cnt = 0;
    for (int i = 1; i <= n; i++) {
      sum[0] = f[i - 1][0];
      for (int j = 1; j <= m / d; j++) sum[j] = sum[j - 1] + f[i - 1][j];
      for (int j = (l[i] + d - 1) / d; j <= m / d; j++) {
        if (j - r[i] / d - 1 < 0)
          f[i][j] = sum[j - (l[i] + d - 1) / d];
        else
          f[i][j] = sum[j - (l[i] + d - 1) / d] - sum[j - r[i] / d - 1];
        f[i][j] %= mod;
      }
    }
    for (int j = 1; j <= m / d; j++) cnt += f[n][j];
    ans += cnt * mu[d];
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 long long N = 1e5 + 10, mod = 998244353;
long long prim[N], nump, mu[N], vis[N], l[N], r[N];
long long f[55][N], sum[N];
int n, m;
void init() {
  mu[1] = 1;
  for (int i = 2; i <= N - 1; i++) {
    if (!vis[i]) {
      prim[++nump] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= nump && i * prim[j] <= N - 1; j++) {
      vis[i * prim[j]] = 1;
      if (i % prim[j] == 0)
        break;
      else
        mu[i * prim[j]] = -mu[i];
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  init();
  for (int i = 1; i <= n; i++) scanf("%d%d", l + i, r + i);
  long long ans = 0;
  for (int d = 1; d <= m; d++) {
    f[0][0] = 1;
    long long cnt = 0;
    for (int i = 1; i <= n; i++) {
      sum[0] = f[i - 1][0];
      for (int j = 1; j <= m / d; j++) sum[j] = sum[j - 1] + f[i - 1][j];
      for (int j = (l[i] + d - 1) / d; j <= m / d; j++) {
        if (j - r[i] / d - 1 < 0)
          f[i][j] = sum[j - (l[i] + d - 1) / d];
        else
          f[i][j] = sum[j - (l[i] + d - 1) / d] - sum[j - r[i] / d - 1];
        f[i][j] %= mod;
      }
    }
    for (int j = 1; j <= m / d; j++) cnt += f[n][j];
    ans += cnt * mu[d];
    ans %= mod;
    ans += mod;
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}

```