#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(51);
const long long MOD = 998244353;
long long mul(long long a, long long b) { return (a * b % MOD); }
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
long long add(long long a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, long long>> v(n);
  vector<long long> ans(m + 1);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
  }
  for (long long i = m; i >= 1; --i) {
    long long val = (m / i + 1);
    vector<long long> dp(val);
    dp[0] = 1;
    for (long long j = 0; j < n; ++j) {
      long long first = ((v[j].first + i - 1) / i), brk = v[j].second / i;
      vector<long long> dpp(val);
      for (long long k = 1; k < val; k++) {
        dp[k] = add(dp[k], dp[k - 1]);
      }
      for (long long f = 0; f < val; f++) {
        if (f >= first) {
          dpp[f] = dp[f - first];
        }
        if (f - brk - 1 >= 0) {
          dpp[f] = sub(dpp[f], dp[f - brk - 1]);
        }
      }
      dp = dpp;
    }
    for (long long f = i; f <= m; f += i) {
      ans[i] = sub(ans[i], ans[f]);
    }
    for (long long j = 0; j < val; j++) {
      ans[i] = add(ans[i], dp[j]);
    }
  }
  cout << ans[1] << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution to 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;
mt19937 rnd(51);
const long long MOD = 998244353;
long long mul(long long a, long long b) { return (a * b % MOD); }
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
long long add(long long a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
  return a;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, long long>> v(n);
  vector<long long> ans(m + 1);
  for (long long i = 0; i < n; i++) {
    cin >> v[i].first >> v[i].second;
  }
  for (long long i = m; i >= 1; --i) {
    long long val = (m / i + 1);
    vector<long long> dp(val);
    dp[0] = 1;
    for (long long j = 0; j < n; ++j) {
      long long first = ((v[j].first + i - 1) / i), brk = v[j].second / i;
      vector<long long> dpp(val);
      for (long long k = 1; k < val; k++) {
        dp[k] = add(dp[k], dp[k - 1]);
      }
      for (long long f = 0; f < val; f++) {
        if (f >= first) {
          dpp[f] = dp[f - first];
        }
        if (f - brk - 1 >= 0) {
          dpp[f] = sub(dpp[f], dp[f - brk - 1]);
        }
      }
      dp = dpp;
    }
    for (long long f = i; f <= m; f += i) {
      ans[i] = sub(ans[i], ans[f]);
    }
    for (long long j = 0; j < val; j++) {
      ans[i] = add(ans[i], dp[j]);
    }
  }
  cout << ans[1] << endl;
  return 0;
}

```