#include <bits/stdc++.h>
int n, m;
int L[55], R[55], x[55], y[55];
long long f[100005], g[100005];
long long dp[100005], pre[100005];
long long solve(int d) {
  for (int i = 1; i <= n; i++) {
    x[i] = (L[i] - 1) / d + 1;
    y[i] = R[i] / d;
  }
  for (int j = 0; j <= m / d; j++) pre[j] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < x[i]; j++) dp[j] = 0;
    for (int j = x[i]; j <= m / d; j++) {
      dp[j] = pre[j - x[i]];
      if (j > y[i]) dp[j] -= pre[j - y[i] - 1];
      dp[j] %= 998244353;
    }
    pre[0] = dp[0];
    for (int j = 1; j <= m / d; j++) pre[j] = (pre[j - 1] + dp[j]) % 998244353;
  }
  return pre[m / d];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d%d", &L[i], &R[i]);
  for (int i = 1; i <= m; i++) f[i] = solve(i);
  for (int i = m; i >= 1; i--) {
    g[i] = f[i];
    for (int j = 2 * i; j <= m; j += i) g[i] -= g[j];
    g[i] %= 998244353;
  }
  printf("%lld\n", (g[1] + 998244353) % 998244353);
  return 0;
}

### Prompt

Generate 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>
int n, m;
int L[55], R[55], x[55], y[55];
long long f[100005], g[100005];
long long dp[100005], pre[100005];
long long solve(int d) {
  for (int i = 1; i <= n; i++) {
    x[i] = (L[i] - 1) / d + 1;
    y[i] = R[i] / d;
  }
  for (int j = 0; j <= m / d; j++) pre[j] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < x[i]; j++) dp[j] = 0;
    for (int j = x[i]; j <= m / d; j++) {
      dp[j] = pre[j - x[i]];
      if (j > y[i]) dp[j] -= pre[j - y[i] - 1];
      dp[j] %= 998244353;
    }
    pre[0] = dp[0];
    for (int j = 1; j <= m / d; j++) pre[j] = (pre[j - 1] + dp[j]) % 998244353;
  }
  return pre[m / d];
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d%d", &L[i], &R[i]);
  for (int i = 1; i <= m; i++) f[i] = solve(i);
  for (int i = m; i >= 1; i--) {
    g[i] = f[i];
    for (int j = 2 * i; j <= m; j += i) g[i] -= g[j];
    g[i] %= 998244353;
  }
  printf("%lld\n", (g[1] + 998244353) % 998244353);
  return 0;
}

```