#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MAXN = 51;
int n, m;
int l[MAXN], r[MAXN];
const int MAXM = 1e5 + 1;
int mu[MAXM];
const int MOD = 998244353;
void add(int& A, int B) {
  A += B;
  if (A >= MOD) A -= MOD;
}
int dp[MAXN][100000 + 1];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &l[i], &r[i]);
  int ans = 0;
  for (int i = 1; i <= 1e5; ++i) {
    int now = i;
    mu[i] = 1;
    for (int j = 2; j * j <= now; j++) {
      if (now % j == 0) {
        if (now % (j * j) == 0) {
          mu[i] = 0;
          break;
        }
        mu[i] *= -1;
        now /= j;
      }
    }
    if (now != 1) mu[i] *= -1;
    mu[i] = (mu[i] + MOD) % MOD;
  }
  dp[0][0] = 1;
  for (int x = 1; x <= m; ++x) {
    if (mu[x]) {
      int M = m / x;
      for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= M; ++j) dp[i][j] = 0;
      for (int i = 1; i <= n; ++i) {
        int l_, r_;
        l_ = (l[i] + x - 1) / x;
        r_ = (r[i]) / x;
        if (l_ <= r_) {
          int sum = 0;
          for (int j = l_; j <= M; ++j) {
            add(sum, dp[i - 1][j - l_]);
            dp[i][j] = sum;
            if (j >= r_) add(sum, MOD - dp[i - 1][j - r_]);
          }
        }
      }
      int sum = 0;
      for (int j = 1; j <= M; ++j) add(sum, dp[n][j]);
      add(ans, 1ll * sum * mu[x] % MOD);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 INF = 0x3f3f3f3f;
const int MAXN = 51;
int n, m;
int l[MAXN], r[MAXN];
const int MAXM = 1e5 + 1;
int mu[MAXM];
const int MOD = 998244353;
void add(int& A, int B) {
  A += B;
  if (A >= MOD) A -= MOD;
}
int dp[MAXN][100000 + 1];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", &l[i], &r[i]);
  int ans = 0;
  for (int i = 1; i <= 1e5; ++i) {
    int now = i;
    mu[i] = 1;
    for (int j = 2; j * j <= now; j++) {
      if (now % j == 0) {
        if (now % (j * j) == 0) {
          mu[i] = 0;
          break;
        }
        mu[i] *= -1;
        now /= j;
      }
    }
    if (now != 1) mu[i] *= -1;
    mu[i] = (mu[i] + MOD) % MOD;
  }
  dp[0][0] = 1;
  for (int x = 1; x <= m; ++x) {
    if (mu[x]) {
      int M = m / x;
      for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= M; ++j) dp[i][j] = 0;
      for (int i = 1; i <= n; ++i) {
        int l_, r_;
        l_ = (l[i] + x - 1) / x;
        r_ = (r[i]) / x;
        if (l_ <= r_) {
          int sum = 0;
          for (int j = l_; j <= M; ++j) {
            add(sum, dp[i - 1][j - l_]);
            dp[i][j] = sum;
            if (j >= r_) add(sum, MOD - dp[i - 1][j - r_]);
          }
        }
      }
      int sum = 0;
      for (int j = 1; j <= M; ++j) add(sum, dp[n][j]);
      add(ans, 1ll * sum * mu[x] % MOD);
    }
  }
  cout << ans << endl;
  return 0;
}

```