#include <bits/stdc++.h>
using namespace std;
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long cdiv(long long a, long long b) {
  return a / b + ((a ^ b) > 0 && a % b);
}
long long fdiv(long long a, long long b) {
  return a / b - ((a ^ b) < 0 && a % b);
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long b, long long e, long long m) {
  b %= m;
  long long ans = 1;
  for (; e; b = b * b % m, e /= 2)
    if (e & 1) ans = ans * b % m;
  return ans;
}
void dbg_out() { cerr << "\n"; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const int N = 100100;
vector<int> pr;
int mu[N];
int mark[N];
void crivo() {
  for (int i = 1; i < (N); ++i) {
    mu[i] = 1;
    mark[i] = 0;
  }
  for (int i = 2; i < (N); ++i) {
    if (!mark[i]) {
      pr.push_back(i);
      mu[i] = -1;
    }
    for (auto p : pr) {
      if (1ll * p * i >= N) break;
      mark[i * p] = 1;
      if (i % p == 0) {
        mu[i * p] = 0;
        break;
      } else
        mu[i * p] = -mu[i];
    }
  }
}
int l[N], r[N];
long long dp[N], pre[N];
const int M = 998244353;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  crivo();
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < (n + 1); ++i) cin >> l[i] >> r[i];
  long long res = 0;
  for (int V = 1; V < (m + 1); ++V)
    if (mu[V] != 0) {
      int mx = m / V;
      for (int i = 0; i < (mx + 1); ++i) pre[i] = dp[i] = 0;
      dp[0] = 1;
      for (int i = 0; i < (mx + 1); ++i) pre[i] = 1;
      int bad = 0;
      for (int k = 1; k < (n + 1); ++k) {
        int L = (l[k] + V - 1) / V;
        int R = r[k] / V;
        if (R == 0 || L > R) {
          bad = 1;
          break;
        }
        for (int i = 0; i < (mx + 1); ++i) {
          if (i - L >= 0)
            dp[i] =
                (M + pre[i - L] - (i - R - 1 >= 0 ? pre[i - R - 1] : 0)) % M;
          else
            dp[i] = 0;
        }
        for (int i = 0; i < (mx + 1); ++i) {
          pre[i] = dp[i] % M;
          if (i) pre[i] = (pre[i] + pre[i - 1]) % M;
        }
      }
      if (!bad) res += mu[V] * pre[mx] % M;
      res %= M;
      if (res < 0) res += M;
    }
  cout << res << "\n";
}

### 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;
std::mt19937 rng(
    (int)std::chrono::steady_clock::now().time_since_epoch().count());
long long cdiv(long long a, long long b) {
  return a / b + ((a ^ b) > 0 && a % b);
}
long long fdiv(long long a, long long b) {
  return a / b - ((a ^ b) < 0 && a % b);
}
long long gcd(long long a, long long b) {
  return (b == 0LL ? a : gcd(b, a % b));
}
long long exp(long long b, long long e, long long m) {
  b %= m;
  long long ans = 1;
  for (; e; b = b * b % m, e /= 2)
    if (e & 1) ans = ans * b % m;
  return ans;
}
void dbg_out() { cerr << "\n"; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const int N = 100100;
vector<int> pr;
int mu[N];
int mark[N];
void crivo() {
  for (int i = 1; i < (N); ++i) {
    mu[i] = 1;
    mark[i] = 0;
  }
  for (int i = 2; i < (N); ++i) {
    if (!mark[i]) {
      pr.push_back(i);
      mu[i] = -1;
    }
    for (auto p : pr) {
      if (1ll * p * i >= N) break;
      mark[i * p] = 1;
      if (i % p == 0) {
        mu[i * p] = 0;
        break;
      } else
        mu[i * p] = -mu[i];
    }
  }
}
int l[N], r[N];
long long dp[N], pre[N];
const int M = 998244353;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  crivo();
  int n, m;
  cin >> n >> m;
  for (int i = 1; i < (n + 1); ++i) cin >> l[i] >> r[i];
  long long res = 0;
  for (int V = 1; V < (m + 1); ++V)
    if (mu[V] != 0) {
      int mx = m / V;
      for (int i = 0; i < (mx + 1); ++i) pre[i] = dp[i] = 0;
      dp[0] = 1;
      for (int i = 0; i < (mx + 1); ++i) pre[i] = 1;
      int bad = 0;
      for (int k = 1; k < (n + 1); ++k) {
        int L = (l[k] + V - 1) / V;
        int R = r[k] / V;
        if (R == 0 || L > R) {
          bad = 1;
          break;
        }
        for (int i = 0; i < (mx + 1); ++i) {
          if (i - L >= 0)
            dp[i] =
                (M + pre[i - L] - (i - R - 1 >= 0 ? pre[i - R - 1] : 0)) % M;
          else
            dp[i] = 0;
        }
        for (int i = 0; i < (mx + 1); ++i) {
          pre[i] = dp[i] % M;
          if (i) pre[i] = (pre[i] + pre[i - 1]) % M;
        }
      }
      if (!bad) res += mu[V] * pre[mx] % M;
      res %= M;
      if (res < 0) res += M;
    }
  cout << res << "\n";
}

```