#include <bits/stdc++.h>
const int MOD = 998244353;
const int MAXN = 52;
const int MAXM = 100002;
int n, m;
int l[MAXN], r[MAXN];
int L[MAXN], R[MAXN];
int mu[MAXM];
int f[MAXN][MAXM];
void Sieve() {
  static bool pri[MAXM];
  static std::vector<int> pml;
  mu[1] = 1;
  for (int i = 2; i < MAXM; ++i) {
    if (!pri[i]) {
      pml.push_back(i);
      mu[i] = -1;
    }
    for (auto &p : pml) {
      if (p * i >= MAXM) break;
      pri[p * i] = true;
      if (i % p == 0) {
        mu[p * i] = 0;
        break;
      }
      mu[p * i] = -mu[i];
    }
  }
}
int Work(int d) {
  for (int i = 1; i <= n; ++i) {
    L[i] = (l[i] + d - 1) / d;
    R[i] = r[i] / d;
    if (L[i] > R[i]) return 0;
  }
  f[0][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= m / d; ++j) f[i][j] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j + L[i] <= m / d; ++j) {
      int st = j + L[i], ed = std::min(j + R[i] + 1, m / d + 1);
      f[i][st] = (f[i][st] + f[i - 1][j]) % MOD;
      f[i][ed] = (f[i][ed] - f[i - 1][j]) % MOD;
    }
    for (int j = 1; j <= m / d; ++j) f[i][j] = (f[i][j] + f[i][j - 1]) % MOD;
  }
  int ret = 0;
  for (int i = 0; i <= m / d; ++i) ret = (ret + f[n][i]) % MOD;
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &l[i], &r[i]);
  Sieve();
  int ans = 0;
  for (int d = 1; d <= m; ++d) ans = (ans + mu[d] * Work(d)) % MOD;
  printf("%d\n", ans < 0 ? ans + MOD : ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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>
const int MOD = 998244353;
const int MAXN = 52;
const int MAXM = 100002;
int n, m;
int l[MAXN], r[MAXN];
int L[MAXN], R[MAXN];
int mu[MAXM];
int f[MAXN][MAXM];
void Sieve() {
  static bool pri[MAXM];
  static std::vector<int> pml;
  mu[1] = 1;
  for (int i = 2; i < MAXM; ++i) {
    if (!pri[i]) {
      pml.push_back(i);
      mu[i] = -1;
    }
    for (auto &p : pml) {
      if (p * i >= MAXM) break;
      pri[p * i] = true;
      if (i % p == 0) {
        mu[p * i] = 0;
        break;
      }
      mu[p * i] = -mu[i];
    }
  }
}
int Work(int d) {
  for (int i = 1; i <= n; ++i) {
    L[i] = (l[i] + d - 1) / d;
    R[i] = r[i] / d;
    if (L[i] > R[i]) return 0;
  }
  f[0][0] = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= m / d; ++j) f[i][j] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j + L[i] <= m / d; ++j) {
      int st = j + L[i], ed = std::min(j + R[i] + 1, m / d + 1);
      f[i][st] = (f[i][st] + f[i - 1][j]) % MOD;
      f[i][ed] = (f[i][ed] - f[i - 1][j]) % MOD;
    }
    for (int j = 1; j <= m / d; ++j) f[i][j] = (f[i][j] + f[i][j - 1]) % MOD;
  }
  int ret = 0;
  for (int i = 0; i <= m / d; ++i) ret = (ret + f[n][i]) % MOD;
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &l[i], &r[i]);
  Sieve();
  int ans = 0;
  for (int d = 1; d <= m; ++d) ans = (ans + mu[d] * Work(d)) % MOD;
  printf("%d\n", ans < 0 ? ans + MOD : ans);
  return 0;
}

```