#include <bits/stdc++.h>
using namespace std;
int n, m;
int l[55], r[55];
long long pie[100100];
long long dp[55][100100];
bool prime[100100];
long long mobius[100100], nummults[100100];
vector<long long> factors[100100];
void comp() {
  for (int i = 1; i < 100100; i++) nummults[i] = 0, mobius[i] = 6969;
  for (int i = 1; i < 100100; i++)
    for (int j = i; j < 100100; j += i) factors[j].push_back(i);
  memset(prime, true, sizeof(prime));
  for (int i = 2; i < 100100; i++) {
    if (!prime[i]) continue;
    for (int j = i * 2; j < 100100; j += i) prime[j] = false;
  }
  mobius[1] = 1;
  mobius[2] = -1;
  for (int i = 3; i < 100100; i++) {
    if (prime[i]) {
      mobius[i] = -1;
      continue;
    }
    for (int j : factors[i]) {
      if (j == 1 || j == i) continue;
      if (!prime[j]) continue;
      int c = i / j;
      if (c % j == 0)
        mobius[i] = 0;
      else
        mobius[i] = mobius[c] * mobius[j];
      break;
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d%d", &l[i], &r[i]);
  for (int p = 1; p <= m; p++) {
    int bound = (m / p);
    for (int i = 0; i <= bound; i++) {
      long long cv = i * p;
      if (cv >= l[0] && cv <= r[0])
        dp[0][i] = 1;
      else
        dp[0][i] = 0;
    }
    for (int j = 1; j < n; j++) {
      long long slide = 0;
      long long lv = l[j] / p;
      if (l[j] % p) lv++;
      int rv = r[j] / p;
      for (int i = 0; i <= bound; i++) {
        if (i - lv >= 0) slide += dp[j - 1][i - lv];
        if (i - rv > 0) slide -= dp[j - 1][i - rv - 1];
        slide += 998244353;
        slide %= 998244353;
        dp[j][i] = slide;
      }
    }
    pie[p] = 0;
    for (int i = 0; i <= bound; i++) pie[p] += dp[n - 1][i];
    pie[p] %= 998244353;
  }
  comp();
  long long ans = 0;
  for (int i = 1; i <= m; i++) {
    ans += mobius[i] * pie[i];
    ans += 998244353;
    ans %= 998244353;
  }
  printf("%lld\n", ans);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int n, m;
int l[55], r[55];
long long pie[100100];
long long dp[55][100100];
bool prime[100100];
long long mobius[100100], nummults[100100];
vector<long long> factors[100100];
void comp() {
  for (int i = 1; i < 100100; i++) nummults[i] = 0, mobius[i] = 6969;
  for (int i = 1; i < 100100; i++)
    for (int j = i; j < 100100; j += i) factors[j].push_back(i);
  memset(prime, true, sizeof(prime));
  for (int i = 2; i < 100100; i++) {
    if (!prime[i]) continue;
    for (int j = i * 2; j < 100100; j += i) prime[j] = false;
  }
  mobius[1] = 1;
  mobius[2] = -1;
  for (int i = 3; i < 100100; i++) {
    if (prime[i]) {
      mobius[i] = -1;
      continue;
    }
    for (int j : factors[i]) {
      if (j == 1 || j == i) continue;
      if (!prime[j]) continue;
      int c = i / j;
      if (c % j == 0)
        mobius[i] = 0;
      else
        mobius[i] = mobius[c] * mobius[j];
      break;
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d%d", &l[i], &r[i]);
  for (int p = 1; p <= m; p++) {
    int bound = (m / p);
    for (int i = 0; i <= bound; i++) {
      long long cv = i * p;
      if (cv >= l[0] && cv <= r[0])
        dp[0][i] = 1;
      else
        dp[0][i] = 0;
    }
    for (int j = 1; j < n; j++) {
      long long slide = 0;
      long long lv = l[j] / p;
      if (l[j] % p) lv++;
      int rv = r[j] / p;
      for (int i = 0; i <= bound; i++) {
        if (i - lv >= 0) slide += dp[j - 1][i - lv];
        if (i - rv > 0) slide -= dp[j - 1][i - rv - 1];
        slide += 998244353;
        slide %= 998244353;
        dp[j][i] = slide;
      }
    }
    pie[p] = 0;
    for (int i = 0; i <= bound; i++) pie[p] += dp[n - 1][i];
    pie[p] %= 998244353;
  }
  comp();
  long long ans = 0;
  for (int i = 1; i <= m; i++) {
    ans += mobius[i] * pie[i];
    ans += 998244353;
    ans %= 998244353;
  }
  printf("%lld\n", ans);
}

```