#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int inf = 1e9;
int mod = 998244353;
inline void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
inline void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
inline int mul(int a, int b) { return (int)((long long)a * b % mod); }
inline int power(int a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline int inv(int a) {
  a %= mod;
  if (a < 0) a += mod;
  int b = mod, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += mod;
  return u;
}
vector<int> dp[2][N];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  int l[n + 1], r[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
  }
  for (int i = 0; i < 2; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j].resize(m / j + 1, 0);
    }
  }
  for (int D = m; D >= 1; D--) {
    int last = 0;
    dp[0][D][0] = 1;
    for (int i = 1; i <= m / D; i++) add(dp[0][D][i], dp[0][D][i - 1]);
    for (int i = 1; i <= n; i++) {
      last ^= 1;
      int y = (l[i] + D - 1) / D;
      int z = r[i] / D;
      if (y > z) {
        for (int j = 0; j <= m / D; j++) {
          dp[last ^ 1][D][j] = 0;
        }
        continue;
      }
      for (int j = 0; j <= m / D; j++) {
        if (j - y >= 0) add(dp[last][D][j], dp[last ^ 1][D][j - y]);
        if (j - z > 0) sub(dp[last][D][j], dp[last ^ 1][D][j - z - 1]);
      }
      for (int j = 0; j <= m / D; j++) {
        if (j) add(dp[last][D][j], dp[last][D][j - 1]);
        dp[last ^ 1][D][j] = 0;
      }
    }
  }
  vector<int> ans(m + 1);
  for (int i = 1; i <= m; i++) ans[i] = dp[n & 1][i][m / i];
  for (int i = m; i > 0; i--) {
    for (int j = i + i; j <= m; j += i) {
      sub(ans[i], ans[j]);
    }
  }
  cout << ans[1] << "\n";
  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;
const int N = 100010;
int inf = 1e9;
int mod = 998244353;
inline void add(int &a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}
inline void sub(int &a, int b) {
  a -= b;
  if (a < 0) a += mod;
}
inline int mul(int a, int b) { return (int)((long long)a * b % mod); }
inline int power(int a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = mul(res, a);
    }
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
inline int inv(int a) {
  a %= mod;
  if (a < 0) a += mod;
  int b = mod, u = 0, v = 1;
  while (a) {
    int t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += mod;
  return u;
}
vector<int> dp[2][N];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  int l[n + 1], r[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> l[i] >> r[i];
  }
  for (int i = 0; i < 2; i++) {
    for (int j = 1; j <= m; j++) {
      dp[i][j].resize(m / j + 1, 0);
    }
  }
  for (int D = m; D >= 1; D--) {
    int last = 0;
    dp[0][D][0] = 1;
    for (int i = 1; i <= m / D; i++) add(dp[0][D][i], dp[0][D][i - 1]);
    for (int i = 1; i <= n; i++) {
      last ^= 1;
      int y = (l[i] + D - 1) / D;
      int z = r[i] / D;
      if (y > z) {
        for (int j = 0; j <= m / D; j++) {
          dp[last ^ 1][D][j] = 0;
        }
        continue;
      }
      for (int j = 0; j <= m / D; j++) {
        if (j - y >= 0) add(dp[last][D][j], dp[last ^ 1][D][j - y]);
        if (j - z > 0) sub(dp[last][D][j], dp[last ^ 1][D][j - z - 1]);
      }
      for (int j = 0; j <= m / D; j++) {
        if (j) add(dp[last][D][j], dp[last][D][j - 1]);
        dp[last ^ 1][D][j] = 0;
      }
    }
  }
  vector<int> ans(m + 1);
  for (int i = 1; i <= m; i++) ans[i] = dp[n & 1][i][m / i];
  for (int i = m; i > 0; i--) {
    for (int j = i + i; j <= m; j += i) {
      sub(ans[i], ans[j]);
    }
  }
  cout << ans[1] << "\n";
  return 0;
}

```