#include <bits/stdc++.h>
using namespace std;
const int maxn = 110000;
const long long modi = 998244353;
long long f[maxn], fact[maxn], dp[50][maxn], pre[50][maxn];
int l[maxn], r[maxn], l2[maxn], r2[maxn], mu[maxn];
bool prime[maxn];
vector<int> p;
long long quickpow(long long base, long long times) {
  long long ans = 1;
  while (times) {
    if (times & 1) ans = ans * base % modi;
    base = base * base % modi;
    times >>= 1;
  }
  return ans;
}
long long inv(long long x) { return quickpow(x, modi - 2); }
int upper_div(int x, int d) {
  if (x % d == 0) return x / d;
  return x / d + 1;
}
inline long long get_pre(int i, int l, int r) {
  if (l > 0) return ((pre[i][r] - pre[i][l - 1]) % modi + modi) % modi;
  return pre[i][r];
}
long long tot(int n, int m) {
  for (int i = 0; i <= m; i++) {
    for (int j = 0; j <= n; j++) {
      dp[j][i] = pre[j][i] = 0;
    }
  }
  dp[0][0] = 1;
  for (int i = 0; i <= m; i++) {
    pre[0][i] = 1;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= m; j++) {
      dp[i][j] = get_pre(i - 1, j - r2[i], j);
    }
    pre[i][0] = dp[i][0];
    for (int j = 1; j <= m; j++) {
      pre[i][j] = (pre[i][j - 1] + dp[i][j]) % modi;
    }
  }
  return pre[n][m];
}
int main() {
  int n, m;
  cin >> n >> m;
  fact[0] = 1;
  for (int i = 1; i <= n + m; i++) {
    fact[i] = fact[i - 1] * i % modi;
  }
  mu[1] = 1;
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
  }
  for (int i = 2; i <= m; i++) {
    if (prime[i] == false) {
      mu[i] = -1;
      p.push_back(i);
    }
    for (auto v : p) {
      if (i * v > m) break;
      mu[i * v] = mu[i] * (-1);
      prime[i * v] = true;
      if (i % v == 0) {
        mu[i * v] = 0;
        break;
      }
    }
  }
  long long ret = 0;
  for (int d = 1; d <= m; d++) {
    int m2 = m / d;
    bool tf = true;
    for (int i = 1; i <= n; i++) {
      l2[i] = upper_div(l[i], d);
      m2 -= l2[i];
      r2[i] = r[i] / d;
      if (l2[i] > r2[i]) {
        tf = false;
        break;
      }
      r2[i] -= l2[i];
    }
    if (tf && m2 >= 0) {
      ret = (ret + mu[d] * tot(n, m2)) % modi;
      ret = (ret + modi) % modi;
    }
  }
  cout << ret << endl;
}

### Prompt

Your task is to create a cpp solution 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;
const int maxn = 110000;
const long long modi = 998244353;
long long f[maxn], fact[maxn], dp[50][maxn], pre[50][maxn];
int l[maxn], r[maxn], l2[maxn], r2[maxn], mu[maxn];
bool prime[maxn];
vector<int> p;
long long quickpow(long long base, long long times) {
  long long ans = 1;
  while (times) {
    if (times & 1) ans = ans * base % modi;
    base = base * base % modi;
    times >>= 1;
  }
  return ans;
}
long long inv(long long x) { return quickpow(x, modi - 2); }
int upper_div(int x, int d) {
  if (x % d == 0) return x / d;
  return x / d + 1;
}
inline long long get_pre(int i, int l, int r) {
  if (l > 0) return ((pre[i][r] - pre[i][l - 1]) % modi + modi) % modi;
  return pre[i][r];
}
long long tot(int n, int m) {
  for (int i = 0; i <= m; i++) {
    for (int j = 0; j <= n; j++) {
      dp[j][i] = pre[j][i] = 0;
    }
  }
  dp[0][0] = 1;
  for (int i = 0; i <= m; i++) {
    pre[0][i] = 1;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= m; j++) {
      dp[i][j] = get_pre(i - 1, j - r2[i], j);
    }
    pre[i][0] = dp[i][0];
    for (int j = 1; j <= m; j++) {
      pre[i][j] = (pre[i][j - 1] + dp[i][j]) % modi;
    }
  }
  return pre[n][m];
}
int main() {
  int n, m;
  cin >> n >> m;
  fact[0] = 1;
  for (int i = 1; i <= n + m; i++) {
    fact[i] = fact[i - 1] * i % modi;
  }
  mu[1] = 1;
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
  }
  for (int i = 2; i <= m; i++) {
    if (prime[i] == false) {
      mu[i] = -1;
      p.push_back(i);
    }
    for (auto v : p) {
      if (i * v > m) break;
      mu[i * v] = mu[i] * (-1);
      prime[i * v] = true;
      if (i % v == 0) {
        mu[i * v] = 0;
        break;
      }
    }
  }
  long long ret = 0;
  for (int d = 1; d <= m; d++) {
    int m2 = m / d;
    bool tf = true;
    for (int i = 1; i <= n; i++) {
      l2[i] = upper_div(l[i], d);
      m2 -= l2[i];
      r2[i] = r[i] / d;
      if (l2[i] > r2[i]) {
        tf = false;
        break;
      }
      r2[i] -= l2[i];
    }
    if (tf && m2 >= 0) {
      ret = (ret + mu[d] * tot(n, m2)) % modi;
      ret = (ret + modi) % modi;
    }
  }
  cout << ret << endl;
}

```