#include <bits/stdc++.h>
using int32 = int;
using int64 = long long;
using i3 = int32;
using i4 = int64;
using pi33 = std::pair<i3, i3>;
using pi34 = std::pair<i3, i4>;
using pi43 = std::pair<i4, i3>;
using pi44 = std::pair<i4, i4>;
const int B = 2;
const int N = 1000005;
const int P = 998244353;
int n, m, d, x, y, z, D, M, c;
int l[N], r[N], p[N], vis[N], e[N];
int64 ans, sum;
int64 dp[N][B];
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0), std::cout.tie(0);
  std::cin >> n >> m;
  for (int i = 2; i <= m; i++) {
    if (!vis[i]) p[++p[0]] = i;
    for (int j = 1; j <= p[0] && i * p[j] <= m; j++) {
      vis[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
  for (int i = 1; i <= n; ++i) std::cin >> l[i] >> r[i];
  e[1] = 1;
  for (int i = 1; i <= m; ++i) {
    if (!vis[i]) {
      e[i] = i == 1 ? 1 : -1;
    } else {
      if (!e[i / vis[i]] || i / vis[i] % vis[i] == 0) continue;
      e[i] = -e[i / vis[i]];
    }
    M = m / i, x = i;
    for (int i = 1; i <= n; ++i) M -= (l[i] - 1) / x;
    for (int i = 0; i <= M; ++i) dp[i][c] = 0;
    dp[0][c] = 1;
    for (int i = 1; i <= n; ++i) {
      D = r[i] / x - (l[i] - 1) / x;
      for (int k = 1; k <= M; ++k) {
        dp[k][c] = (dp[k - 1][c] + dp[k][c]) % P;
      }
      for (int k = 0; k <= M; ++k) {
        dp[k][!c] = dp[k - 1][c];
        if (k > D) dp[k][!c] -= dp[k - D - 1][c];
      }
      c ^= 1;
    }
    sum = 0;
    for (int i = 0; i <= M; ++i) sum += dp[i][c];
    ans += (sum * e[i] % P + P) % P;
  }
  std::cout << ans % P << std::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 int32 = int;
using int64 = long long;
using i3 = int32;
using i4 = int64;
using pi33 = std::pair<i3, i3>;
using pi34 = std::pair<i3, i4>;
using pi43 = std::pair<i4, i3>;
using pi44 = std::pair<i4, i4>;
const int B = 2;
const int N = 1000005;
const int P = 998244353;
int n, m, d, x, y, z, D, M, c;
int l[N], r[N], p[N], vis[N], e[N];
int64 ans, sum;
int64 dp[N][B];
int main() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0), std::cout.tie(0);
  std::cin >> n >> m;
  for (int i = 2; i <= m; i++) {
    if (!vis[i]) p[++p[0]] = i;
    for (int j = 1; j <= p[0] && i * p[j] <= m; j++) {
      vis[i * p[j]] = p[j];
      if (i % p[j] == 0) break;
    }
  }
  for (int i = 1; i <= n; ++i) std::cin >> l[i] >> r[i];
  e[1] = 1;
  for (int i = 1; i <= m; ++i) {
    if (!vis[i]) {
      e[i] = i == 1 ? 1 : -1;
    } else {
      if (!e[i / vis[i]] || i / vis[i] % vis[i] == 0) continue;
      e[i] = -e[i / vis[i]];
    }
    M = m / i, x = i;
    for (int i = 1; i <= n; ++i) M -= (l[i] - 1) / x;
    for (int i = 0; i <= M; ++i) dp[i][c] = 0;
    dp[0][c] = 1;
    for (int i = 1; i <= n; ++i) {
      D = r[i] / x - (l[i] - 1) / x;
      for (int k = 1; k <= M; ++k) {
        dp[k][c] = (dp[k - 1][c] + dp[k][c]) % P;
      }
      for (int k = 0; k <= M; ++k) {
        dp[k][!c] = dp[k - 1][c];
        if (k > D) dp[k][!c] -= dp[k - D - 1][c];
      }
      c ^= 1;
    }
    sum = 0;
    for (int i = 0; i <= M; ++i) sum += dp[i][c];
    ans += (sum * e[i] % P + P) % P;
  }
  std::cout << ans % P << std::endl;
  return 0;
}

```