#include <bits/stdc++.h>
using namespace std;
const int max_n = 50 + 5;
int l[max_n], r[max_n], L[max_n], R[max_n];
const int max_m = 1e5 + 5;
int dp[max_n][max_m], pre[max_n][max_m], f[max_m];
const int mod = 998244353;
inline void add(int &a, int b) { a = a + b - (a + b >= mod ? mod : 0); }
inline void sub(int &a, int b) { a = a - b + (a < b ? mod : 0); }
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", l + i, r + i);
  for (int d = 1; d <= m; ++d) {
    bool flag_break = false;
    int sL = 0;
    for (int i = 1; i <= n; ++i) {
      L[i] = (l[i] + d - 1) / d;
      R[i] = r[i] / d;
      if (L[i] > R[i]) {
        flag_break = true;
        break;
      }
      sL += L[i];
    }
    if (flag_break) continue;
    int M = m / d;
    if (M < sL) continue;
    dp[0][0] = 1;
    for (int i = 0; i <= M; ++i) pre[0][i] = 1;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= M; ++j) {
        int l = j - R[i], r = j - L[i];
        if (r >= 0) dp[i][j] = pre[i - 1][r];
        if (l >= 1) sub(dp[i][j], pre[i - 1][l - 1]);
        pre[i][j] = pre[i][j - 1];
        add(pre[i][j], dp[i][j]);
      }
    f[d] = pre[n][M];
  }
  for (int i = m; i >= 1; --i)
    for (int j = i << 1; j <= m; j += i) sub(f[i], f[j]);
  printf("%d\n", f[1]);
  return 0;
}

### 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;
const int max_n = 50 + 5;
int l[max_n], r[max_n], L[max_n], R[max_n];
const int max_m = 1e5 + 5;
int dp[max_n][max_m], pre[max_n][max_m], f[max_m];
const int mod = 998244353;
inline void add(int &a, int b) { a = a + b - (a + b >= mod ? mod : 0); }
inline void sub(int &a, int b) { a = a - b + (a < b ? mod : 0); }
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d%d", l + i, r + i);
  for (int d = 1; d <= m; ++d) {
    bool flag_break = false;
    int sL = 0;
    for (int i = 1; i <= n; ++i) {
      L[i] = (l[i] + d - 1) / d;
      R[i] = r[i] / d;
      if (L[i] > R[i]) {
        flag_break = true;
        break;
      }
      sL += L[i];
    }
    if (flag_break) continue;
    int M = m / d;
    if (M < sL) continue;
    dp[0][0] = 1;
    for (int i = 0; i <= M; ++i) pre[0][i] = 1;
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= M; ++j) {
        int l = j - R[i], r = j - L[i];
        if (r >= 0) dp[i][j] = pre[i - 1][r];
        if (l >= 1) sub(dp[i][j], pre[i - 1][l - 1]);
        pre[i][j] = pre[i][j - 1];
        add(pre[i][j], dp[i][j]);
      }
    f[d] = pre[n][M];
  }
  for (int i = m; i >= 1; --i)
    for (int j = i << 1; j <= m; j += i) sub(f[i], f[j]);
  printf("%d\n", f[1]);
  return 0;
}

```