#include <bits/stdc++.h>
const int INF = 2e9;
inline int Max(const int &a, const int &b) { return a > b ? a : b; }
inline int Min(const int &a, const int &b) { return a > b ? b : a; }
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int N = 55;
const int M = 1e5 + 7;
const int mod = 998244353;
int n, m;
int l[N], r[N];
int mu[M];
int prime[M], ntp[M], cnt;
void xxs(int n) {
  mu[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!ntp[i]) {
      prime[++cnt] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= cnt; ++j) {
      if (i * prime[j] > n) break;
      ntp[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}
long long dp[M], s[M], ans;
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) l[i] = read(), r[i] = read();
  xxs(m);
  for (int i = 1; i <= m; ++i) {
    if (mu[i] == 0) continue;
    int lim = m / i;
    for (int j = 1; j <= lim; ++j) dp[j] = 0;
    dp[0] = 1;
    for (int j = 1; j <= n; ++j) {
      int L = (l[j] + i - 1) / i, R = r[j] / i;
      for (int k = 0; k <= lim; ++k) {
        s[k] = (k > 0 ? s[k - 1] : 0);
        s[k] = (s[k] + dp[k]) % mod;
      }
      for (int k = 0; k <= lim; ++k) {
        dp[k] = (k - L >= 0 ? s[k - L] : 0);
        dp[k] = (dp[k] - (k - R - 1 >= 0 ? s[k - R - 1] : 0) + mod) % mod;
      }
    }
    long long res = 0;
    for (int k = 0; k <= lim; ++k) res = (res + dp[k]) % mod;
    ans = (ans + res * mu[i] + mod) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
const int INF = 2e9;
inline int Max(const int &a, const int &b) { return a > b ? a : b; }
inline int Min(const int &a, const int &b) { return a > b ? b : a; }
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const int N = 55;
const int M = 1e5 + 7;
const int mod = 998244353;
int n, m;
int l[N], r[N];
int mu[M];
int prime[M], ntp[M], cnt;
void xxs(int n) {
  mu[1] = 1;
  for (int i = 2; i <= n; ++i) {
    if (!ntp[i]) {
      prime[++cnt] = i;
      mu[i] = -1;
    }
    for (int j = 1; j <= cnt; ++j) {
      if (i * prime[j] > n) break;
      ntp[i * prime[j]] = 1;
      if (i % prime[j] == 0) {
        mu[i * prime[j]] = 0;
        break;
      }
      mu[i * prime[j]] = -mu[i];
    }
  }
}
long long dp[M], s[M], ans;
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; ++i) l[i] = read(), r[i] = read();
  xxs(m);
  for (int i = 1; i <= m; ++i) {
    if (mu[i] == 0) continue;
    int lim = m / i;
    for (int j = 1; j <= lim; ++j) dp[j] = 0;
    dp[0] = 1;
    for (int j = 1; j <= n; ++j) {
      int L = (l[j] + i - 1) / i, R = r[j] / i;
      for (int k = 0; k <= lim; ++k) {
        s[k] = (k > 0 ? s[k - 1] : 0);
        s[k] = (s[k] + dp[k]) % mod;
      }
      for (int k = 0; k <= lim; ++k) {
        dp[k] = (k - L >= 0 ? s[k - L] : 0);
        dp[k] = (dp[k] - (k - R - 1 >= 0 ? s[k - R - 1] : 0) + mod) % mod;
      }
    }
    long long res = 0;
    for (int k = 0; k <= lim; ++k) res = (res + dp[k]) % mod;
    ans = (ans + res * mu[i] + mod) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

```