#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
int mu[100007], ck[100007], prime[100007], tot;
inline void init() {
  mu[1] = 1;
  for (int i = 2; i < 100007; i++) {
    if (!ck[i]) prime[++tot] = i, mu[i] = -1;
    for (int j = 1; j <= tot && i * prime[j] < 100007; j++) {
      int now = i * prime[j];
      ck[now] = 1;
      if (i % prime[j])
        mu[now] = -mu[i];
      else {
        mu[now] = 0;
        break;
      }
    }
  }
}
int t, n, m;
struct node {
  int l, r;
} a[57], aa[57];
long long sum[100007], dp[55][100007];
long long DP(int d) {
  int mm = m / d;
  for (int i = 1; i <= n; i++) {
    aa[i].l = (a[i].l - 1) / d + 1;
    aa[i].r = a[i].r / d;
  }
  sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    if (aa[i].l > aa[i].r) return 0;
    for (int j = 1; j <= mm; j++) {
      if (i == 1 && j >= aa[i].l && j <= aa[i].r)
        dp[i][j] = 1;
      else if (i == 1)
        dp[i][j] = 0;
      else {
        int le = max(0, j - aa[i].r - 1), ri = max(0, j - aa[i].l);
        dp[i][j] = (sum[ri] - sum[le] + mod) % mod;
      }
    }
    for (int j = 1; j <= mm; j++) sum[j] = (sum[j - 1] + dp[i][j]) % mod;
  }
  return sum[mm];
}
int main() {
  t = 1;
  while (t--) {
    scanf("%d%d", &n, &m);
    init();
    for (int i = 1; i <= n; i++) {
      a[i].l = read(), a[i].r = read();
    }
    long long ans = 0;
    for (int d = 1; d <= m; d++) {
      if (!mu[d]) continue;
      ans = (ans + DP(d) * mu[d] + mod) % mod;
    }
    printf("%lld\n", ans);
  }
}

### 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 inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 998244353;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
int mu[100007], ck[100007], prime[100007], tot;
inline void init() {
  mu[1] = 1;
  for (int i = 2; i < 100007; i++) {
    if (!ck[i]) prime[++tot] = i, mu[i] = -1;
    for (int j = 1; j <= tot && i * prime[j] < 100007; j++) {
      int now = i * prime[j];
      ck[now] = 1;
      if (i % prime[j])
        mu[now] = -mu[i];
      else {
        mu[now] = 0;
        break;
      }
    }
  }
}
int t, n, m;
struct node {
  int l, r;
} a[57], aa[57];
long long sum[100007], dp[55][100007];
long long DP(int d) {
  int mm = m / d;
  for (int i = 1; i <= n; i++) {
    aa[i].l = (a[i].l - 1) / d + 1;
    aa[i].r = a[i].r / d;
  }
  sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    if (aa[i].l > aa[i].r) return 0;
    for (int j = 1; j <= mm; j++) {
      if (i == 1 && j >= aa[i].l && j <= aa[i].r)
        dp[i][j] = 1;
      else if (i == 1)
        dp[i][j] = 0;
      else {
        int le = max(0, j - aa[i].r - 1), ri = max(0, j - aa[i].l);
        dp[i][j] = (sum[ri] - sum[le] + mod) % mod;
      }
    }
    for (int j = 1; j <= mm; j++) sum[j] = (sum[j - 1] + dp[i][j]) % mod;
  }
  return sum[mm];
}
int main() {
  t = 1;
  while (t--) {
    scanf("%d%d", &n, &m);
    init();
    for (int i = 1; i <= n; i++) {
      a[i].l = read(), a[i].r = read();
    }
    long long ans = 0;
    for (int d = 1; d <= m; d++) {
      if (!mu[d]) continue;
      ans = (ans + DP(d) * mu[d] + mod) % mod;
    }
    printf("%lld\n", ans);
  }
}

```