#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
const long long mod = 998244353;
int cnt;
int mo[N];
bool vis[N];
int prime[N];
struct Node {
  int l, r;
} q[N];
int n, m;
void initPrime(const int& n) {
  mo[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) {
      mo[i] = -1;
      prime[++cnt] = i;
    }
    for (int j = 1; i <= n / prime[j]; j++) {
      vis[i * prime[j]] = true;
      if (i % prime[j] == 0) {
        mo[i * prime[j]] = 0;
        break;
      }
      mo[i * prime[j]] = -mo[i];
    }
  }
}
int d[N];
int pre[N];
long long get(long long g) {
  vector<Node> vt;
  for (int i = 1; i <= n; i++) {
    int l = (q[i].l + g - 1) / g;
    int r = q[i].r / g;
    if (l > r) return 0;
    vt.push_back({l, r});
  }
  int k = m / g;
  for (int i = vt[0].l; i <= vt[0].r; i++) {
    int l = i + vt[1].l;
    int r = min(k, i + vt[1].r);
    if (l > k) break;
    d[l] += 1, d[r + 1] -= 1;
  }
  for (int i = 1; i <= k; i++) {
    pre[i] = pre[i - 1] + d[i];
    d[i] = 0;
  }
  d[k + 1] = 0;
  for (int i = 2; i <= n - 1; i++) {
    for (int j = 1; j <= k; j++) {
      if (pre[j]) {
        int l = j + vt[i].l;
        int r = min(k, j + vt[i].r);
        if (l > k) break;
        d[l] = (d[l] + pre[j]) % mod;
        d[r + 1] = (d[r + 1] - pre[j] + mod) % mod;
      }
    }
    for (int j = 1; j <= k; j++) {
      pre[j] = (pre[j - 1] + d[j]) % mod;
      d[j] = 0;
    }
    d[k + 1] = 0;
  }
  int res = 0;
  for (int i = 1; i <= k; i++) {
    res = (res + pre[i]) % mod;
  }
  return res;
}
int main() {
  initPrime(1e5);
  int mi = 1e5;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> q[i].l >> q[i].r;
    mi = min(mi, q[i].r);
  }
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  int ans = 0;
  for (long long i = 1; i <= mi; i++) {
    if (mo[i] == 0) continue;
    ans = (ans + mo[i] * get(i) + mod) % mod;
  }
  cout << ans << 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;
const int N = 1e5 + 10;
const long long mod = 998244353;
int cnt;
int mo[N];
bool vis[N];
int prime[N];
struct Node {
  int l, r;
} q[N];
int n, m;
void initPrime(const int& n) {
  mo[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (!vis[i]) {
      mo[i] = -1;
      prime[++cnt] = i;
    }
    for (int j = 1; i <= n / prime[j]; j++) {
      vis[i * prime[j]] = true;
      if (i % prime[j] == 0) {
        mo[i * prime[j]] = 0;
        break;
      }
      mo[i * prime[j]] = -mo[i];
    }
  }
}
int d[N];
int pre[N];
long long get(long long g) {
  vector<Node> vt;
  for (int i = 1; i <= n; i++) {
    int l = (q[i].l + g - 1) / g;
    int r = q[i].r / g;
    if (l > r) return 0;
    vt.push_back({l, r});
  }
  int k = m / g;
  for (int i = vt[0].l; i <= vt[0].r; i++) {
    int l = i + vt[1].l;
    int r = min(k, i + vt[1].r);
    if (l > k) break;
    d[l] += 1, d[r + 1] -= 1;
  }
  for (int i = 1; i <= k; i++) {
    pre[i] = pre[i - 1] + d[i];
    d[i] = 0;
  }
  d[k + 1] = 0;
  for (int i = 2; i <= n - 1; i++) {
    for (int j = 1; j <= k; j++) {
      if (pre[j]) {
        int l = j + vt[i].l;
        int r = min(k, j + vt[i].r);
        if (l > k) break;
        d[l] = (d[l] + pre[j]) % mod;
        d[r + 1] = (d[r + 1] - pre[j] + mod) % mod;
      }
    }
    for (int j = 1; j <= k; j++) {
      pre[j] = (pre[j - 1] + d[j]) % mod;
      d[j] = 0;
    }
    d[k + 1] = 0;
  }
  int res = 0;
  for (int i = 1; i <= k; i++) {
    res = (res + pre[i]) % mod;
  }
  return res;
}
int main() {
  initPrime(1e5);
  int mi = 1e5;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> q[i].l >> q[i].r;
    mi = min(mi, q[i].r);
  }
  if (n == 1) {
    cout << 1 << endl;
    return 0;
  }
  int ans = 0;
  for (long long i = 1; i <= mi; i++) {
    if (mo[i] == 0) continue;
    ans = (ans + mo[i] * get(i) + mod) % mod;
  }
  cout << ans << endl;
  return 0;
}

```