#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
const int M = 3e3 + 10;
const int N = 1e5 + 10;
int l[N], r[N];
long long dp[60][N];
long long cnt[N];
long long sum[N];
template <class T>
inline void rd(T& x) {
  char c = getchar();
  long long res = 0, f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    res = (res << 1) + (res << 3) + (c ^ 48);
    c = getchar();
  }
  x = res * f;
}
void run() {
  int n, m;
  rd(n), rd(m);
  for (int i = 1; i <= n; i++) {
    rd(l[i]), rd(r[i]);
  }
  for (int k = 1; k <= m; k++) {
    int top = m / k;
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      int tl = (l[i] + k - 1) / k;
      int tr = r[i] / k;
      sum[0] = dp[i - 1][0];
      for (int j = 1; j <= top; j++) {
        sum[j] = sum[j - 1] + dp[i - 1][j];
        sum[j] %= mod;
        dp[i - 1][j] = 0;
      }
      for (int j = tl; j <= top; j++) {
        dp[i][j] += sum[j - tl] - ((j - tr - 1 < 0) ? 0 : sum[j - tr - 1]);
        dp[i][j] %= mod;
      }
    }
    for (int i = 1; i <= top; i++) {
      cnt[k] += dp[n][i];
      dp[n][i] = 0;
      cnt[k] %= mod;
    }
  }
  for (int i = m; i > 0; i--) {
    for (int j = 2; j * i <= m; j++) {
      cnt[i] -= cnt[i * j];
      cnt[i] += mod;
      cnt[i] %= mod;
    }
  }
  printf("%lld\n", cnt[1]);
}
int main() {
  run();
  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 mod = 998244353;
const int M = 3e3 + 10;
const int N = 1e5 + 10;
int l[N], r[N];
long long dp[60][N];
long long cnt[N];
long long sum[N];
template <class T>
inline void rd(T& x) {
  char c = getchar();
  long long res = 0, f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    res = (res << 1) + (res << 3) + (c ^ 48);
    c = getchar();
  }
  x = res * f;
}
void run() {
  int n, m;
  rd(n), rd(m);
  for (int i = 1; i <= n; i++) {
    rd(l[i]), rd(r[i]);
  }
  for (int k = 1; k <= m; k++) {
    int top = m / k;
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      int tl = (l[i] + k - 1) / k;
      int tr = r[i] / k;
      sum[0] = dp[i - 1][0];
      for (int j = 1; j <= top; j++) {
        sum[j] = sum[j - 1] + dp[i - 1][j];
        sum[j] %= mod;
        dp[i - 1][j] = 0;
      }
      for (int j = tl; j <= top; j++) {
        dp[i][j] += sum[j - tl] - ((j - tr - 1 < 0) ? 0 : sum[j - tr - 1]);
        dp[i][j] %= mod;
      }
    }
    for (int i = 1; i <= top; i++) {
      cnt[k] += dp[n][i];
      dp[n][i] = 0;
      cnt[k] %= mod;
    }
  }
  for (int i = m; i > 0; i--) {
    for (int j = 2; j * i <= m; j++) {
      cnt[i] -= cnt[i * j];
      cnt[i] += mod;
      cnt[i] %= mod;
    }
  }
  printf("%lld\n", cnt[1]);
}
int main() {
  run();
  return 0;
}

```