#include <bits/stdc++.h>
const long long MOD = 998244353;
using pii = std::pair<long long, long long>;
using namespace std;
const long long maxn = 55, maxrange = 1e5 + 5;
long long n, m, l[maxn], r[maxn], dp[2][maxrange], ways[maxrange];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> l[i] >> r[i];
  long long maxgcd = m / n;
  for (long long gcd = maxgcd; gcd >= 1; gcd--) {
    long long compressed_till = m / gcd;
    for (long long j = 0; j <= compressed_till; j++) dp[0][j] = 0;
    dp[0][0] = 1;
    for (long long i = 0; i < n; i++) {
      long long cur = i & 1;
      long long ne = cur ^ 1;
      for (long long j = 0; j <= compressed_till; j++) dp[ne][j] = 0;
      for (long long j = 0; j < compressed_till; j++) {
        if (j > 0) {
          dp[cur][j] += dp[cur][j - 1];
          dp[cur][j] %= MOD;
        }
        long long from = j + (l[i] + gcd - 1) / gcd;
        long long to = min(j + r[i] / gcd, compressed_till);
        if (from > to) continue;
        dp[ne][from] += dp[cur][j];
        dp[ne][from] %= MOD;
        dp[ne][to + 1] += (MOD - dp[cur][j]);
        dp[ne][to + 1] %= MOD;
      }
    }
    for (long long j = 0; j <= compressed_till; j++) {
      ways[gcd] += dp[n & 1][j];
      ways[gcd] %= MOD;
    }
    for (long long mult = 2; gcd * mult <= m; mult++) {
      ways[gcd] += (MOD - ways[gcd * mult]);
      ways[gcd] %= MOD;
    }
  }
  cout << ways[1] << "\n";
  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>
const long long MOD = 998244353;
using pii = std::pair<long long, long long>;
using namespace std;
const long long maxn = 55, maxrange = 1e5 + 5;
long long n, m, l[maxn], r[maxn], dp[2][maxrange], ways[maxrange];
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < n; i++) cin >> l[i] >> r[i];
  long long maxgcd = m / n;
  for (long long gcd = maxgcd; gcd >= 1; gcd--) {
    long long compressed_till = m / gcd;
    for (long long j = 0; j <= compressed_till; j++) dp[0][j] = 0;
    dp[0][0] = 1;
    for (long long i = 0; i < n; i++) {
      long long cur = i & 1;
      long long ne = cur ^ 1;
      for (long long j = 0; j <= compressed_till; j++) dp[ne][j] = 0;
      for (long long j = 0; j < compressed_till; j++) {
        if (j > 0) {
          dp[cur][j] += dp[cur][j - 1];
          dp[cur][j] %= MOD;
        }
        long long from = j + (l[i] + gcd - 1) / gcd;
        long long to = min(j + r[i] / gcd, compressed_till);
        if (from > to) continue;
        dp[ne][from] += dp[cur][j];
        dp[ne][from] %= MOD;
        dp[ne][to + 1] += (MOD - dp[cur][j]);
        dp[ne][to + 1] %= MOD;
      }
    }
    for (long long j = 0; j <= compressed_till; j++) {
      ways[gcd] += dp[n & 1][j];
      ways[gcd] %= MOD;
    }
    for (long long mult = 2; gcd * mult <= m; mult++) {
      ways[gcd] += (MOD - ways[gcd * mult]);
      ways[gcd] %= MOD;
    }
  }
  cout << ways[1] << "\n";
  return 0;
}

```