#include <bits/stdc++.h>
using namespace std;
const int N = 100002, M = 998244353;
int n, m, t, i, j, L[N], R[N], l, r, o[N], f[N], Ans[N];
void Add(int &x, int y) {
  x += y;
  x = x >= M ? x - M : x;
}
int Q(int p) {
  int i, j, k;
  t = m / p;
  for (i = 1; i <= n; i++) {
    r = R[i] / p;
    l = (L[i] + p - 1) / p;
    t -= l;
    o[i] = r - l + 1;
  }
  if (t < 0) return 0;
  for (i = 0; i <= t; i++) f[i] = 0;
  f[0] = 1;
  for (i = 1; i <= n; i++) {
    k = o[i];
    for (j = t; j >= k; j--) Add(f[j], -f[j - k] + M);
  }
  for (i = 0; i <= n; i++)
    for (j = 1; j <= t; j++) Add(f[j], f[j - 1]);
  return f[t];
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d%d", L + i, R + i);
  for (i = 1; i <= m; i++) Ans[i] = Q(i);
  for (i = m; i; i--)
    for (j = i * 2; j <= m; j += i) Add(Ans[i], M - Ans[j]);
  printf("%d\n", Ans[1]);
}

### Prompt

Your challenge is to write 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 = 100002, M = 998244353;
int n, m, t, i, j, L[N], R[N], l, r, o[N], f[N], Ans[N];
void Add(int &x, int y) {
  x += y;
  x = x >= M ? x - M : x;
}
int Q(int p) {
  int i, j, k;
  t = m / p;
  for (i = 1; i <= n; i++) {
    r = R[i] / p;
    l = (L[i] + p - 1) / p;
    t -= l;
    o[i] = r - l + 1;
  }
  if (t < 0) return 0;
  for (i = 0; i <= t; i++) f[i] = 0;
  f[0] = 1;
  for (i = 1; i <= n; i++) {
    k = o[i];
    for (j = t; j >= k; j--) Add(f[j], -f[j - k] + M);
  }
  for (i = 0; i <= n; i++)
    for (j = 1; j <= t; j++) Add(f[j], f[j - 1]);
  return f[t];
}
int main() {
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) scanf("%d%d", L + i, R + i);
  for (i = 1; i <= m; i++) Ans[i] = Q(i);
  for (i = m; i; i--)
    for (j = i * 2; j <= m; j += i) Add(Ans[i], M - Ans[j]);
  printf("%d\n", Ans[1]);
}

```