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.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, mod = 998244353;
long long f[55][N], l[N], r[N], sum[N], prim[N], nump, vis[N], mu[N];
void init() {
  mu[1] = 1;
  for (int i = 2; i < N; i++) {
    if (!vis[i]) {
      prim[++nump] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= nump && i * prim[j] < N; j++) {
      vis[i * prim[j]] = 1;
      if (i % prim[j] == 0)
        break;
      else
        mu[i * prim[j]] = -mu[i];
    }
  }
}
int main() {
  long long n, m;
  scanf("%lld%lld", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld%lld", l + i, r + i);
  long long ans = 0;
  init();
  for (int d = 1; d <= m; d++) {
    f[0][0] = 1;
    long long cnt = 0;
    if (mu[d]) {
      for (int i = 1; i <= n; i++) {
        sum[0] = f[i - 1][0];
        for (int j = 1; j <= m / d; j++)
          sum[j] = (sum[j - 1] + f[i - 1][j]) % mod;
        for (int j = (l[i] + d - 1) / d; j <= m / d; j++) {
          if (j >= r[i] / d)
            f[i][j] = sum[j - (l[i] + d - 1) / d] - sum[j - r[i] / d - 1];
          else
            f[i][j] = sum[j - (l[i] + d - 1) / d];
          f[i][j] = f[i][j] % mod;
        }
      }
      for (int j = 1; j <= m / d; j++) cnt = (cnt + f[n][j]) % mod;
      ans += mu[d] * cnt;
      ans = ans % mod;
      ans += mod;
      ans = ans % mod;
    }
  }
  cout << ans << endl;
  return 0;
}
