#include <bits/stdc++.h>
using namespace std;
const long long MOD = (998244353);
const long long MAX_VALEURS = (200 * 1000);
const long long MAX_ETOILES = (50);
long long Ans[MAX_VALEURS];
long long DP[MAX_VALEURS];
pair<long long, long long> Bornes[MAX_ETOILES];
long long nbEtoiles, borneSup;
long long FindDP(long long d) {
  long long borne = borneSup / d;
  fill_n(DP + 1, borne, 0LL);
  DP[0] = 1;
  for (long long i = 0; i < nbEtoiles; i++) {
    long long sum = 0;
    long long inf = (Bornes[i].first + d - 1) / d, sup = Bornes[i].second / d;
    for (long long j = inf; j <= sup; j++) {
      sum += DP[borne - j];
    }
    sum %= MOD;
    for (long long j = borne; j >= 0; j--) {
      DP[j] = sum;
      if (j >= inf) sum -= DP[j - inf];
      if (j > sup) sum += DP[j - sup - 1];
      sum %= MOD;
      if (sum < 0) sum += MOD;
    }
  }
  long long ans = 0;
  for (long long i = 0; i <= borne; i++) {
    ans += DP[i];
  }
  return ans % MOD;
}
void Solve() {
  scanf("%lld %lld", &nbEtoiles, &borneSup);
  for (long long i = 0; i < nbEtoiles; i++) {
    scanf("%lld %lld", &Bornes[i].first, &Bornes[i].second);
  }
  for (long long i = 1; i <= borneSup; i++) {
    Ans[i] = FindDP(i);
  }
  for (long long i = borneSup; i > 0; i--) {
    for (long long j = 2 * i; j <= borneSup; j += i) {
      Ans[i] -= Ans[j];
    }
    Ans[i] %= MOD;
    if (Ans[i] < 0) Ans[i] += MOD;
  }
  printf("%lld\n", Ans[1]);
  return;
}
signed main() {
  long long nbTests = 1;
  while (nbTests--) {
    Solve();
  }
  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>
using namespace std;
const long long MOD = (998244353);
const long long MAX_VALEURS = (200 * 1000);
const long long MAX_ETOILES = (50);
long long Ans[MAX_VALEURS];
long long DP[MAX_VALEURS];
pair<long long, long long> Bornes[MAX_ETOILES];
long long nbEtoiles, borneSup;
long long FindDP(long long d) {
  long long borne = borneSup / d;
  fill_n(DP + 1, borne, 0LL);
  DP[0] = 1;
  for (long long i = 0; i < nbEtoiles; i++) {
    long long sum = 0;
    long long inf = (Bornes[i].first + d - 1) / d, sup = Bornes[i].second / d;
    for (long long j = inf; j <= sup; j++) {
      sum += DP[borne - j];
    }
    sum %= MOD;
    for (long long j = borne; j >= 0; j--) {
      DP[j] = sum;
      if (j >= inf) sum -= DP[j - inf];
      if (j > sup) sum += DP[j - sup - 1];
      sum %= MOD;
      if (sum < 0) sum += MOD;
    }
  }
  long long ans = 0;
  for (long long i = 0; i <= borne; i++) {
    ans += DP[i];
  }
  return ans % MOD;
}
void Solve() {
  scanf("%lld %lld", &nbEtoiles, &borneSup);
  for (long long i = 0; i < nbEtoiles; i++) {
    scanf("%lld %lld", &Bornes[i].first, &Bornes[i].second);
  }
  for (long long i = 1; i <= borneSup; i++) {
    Ans[i] = FindDP(i);
  }
  for (long long i = borneSup; i > 0; i--) {
    for (long long j = 2 * i; j <= borneSup; j += i) {
      Ans[i] -= Ans[j];
    }
    Ans[i] %= MOD;
    if (Ans[i] < 0) Ans[i] += MOD;
  }
  printf("%lld\n", Ans[1]);
  return;
}
signed main() {
  long long nbTests = 1;
  while (nbTests--) {
    Solve();
  }
  return 0;
}

```