#include <bits/stdc++.h>
using namespace std;
int i, j, k, n, m, t, l[55], r[55], cnt;
int g[100500], f[100500][55], pre[100500][55];
int su(int a, int b) {
  a += b;
  return (a >= 998244353) ? a - 998244353 : a;
}
void get(int x) {
  int i, j, k, res = 0, l1, r1;
  for (i = x; i <= m; i += x) {
    memset(f[i], 0, sizeof(f[i]));
    memset(pre[i], 0, sizeof(pre[i]));
  }
  for (i = 1; i <= n; i++) {
    r1 = (r[i] / x) * x;
    l1 = ((l[i] + x - 1) / x) * x;
    if (l1 > r1) {
      return;
    }
    if (i == 1) {
      for (j = l1; j <= r1; j += x) {
        f[j][i] = 1;
      }
    } else {
      for (j = x; j <= m; j += x) {
        cnt++;
        if (j <= l1) {
          continue;
        }
        if (j <= r1 + x) {
          f[j][i] = pre[j - l1][i - 1];
        } else {
          f[j][i] = su(pre[j - l1][i - 1], 998244353 - pre[j - r1 - x][i - 1]);
        }
      }
    }
    for (j = x; j <= m; j += x) {
      pre[j][i] = su(pre[j - x][i], f[j][i]);
    }
  }
  for (i = x; i <= m; i += x) {
    res = su(res, f[i][n]);
  }
  g[x] = res;
}
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++) {
    get(i);
  }
  for (i = m; i >= 1; i--) {
    for (j = i + i; j <= m; j += i) {
      g[i] = su(g[i], 998244353 - g[j]);
    }
  }
  printf("%d", g[1]);
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int i, j, k, n, m, t, l[55], r[55], cnt;
int g[100500], f[100500][55], pre[100500][55];
int su(int a, int b) {
  a += b;
  return (a >= 998244353) ? a - 998244353 : a;
}
void get(int x) {
  int i, j, k, res = 0, l1, r1;
  for (i = x; i <= m; i += x) {
    memset(f[i], 0, sizeof(f[i]));
    memset(pre[i], 0, sizeof(pre[i]));
  }
  for (i = 1; i <= n; i++) {
    r1 = (r[i] / x) * x;
    l1 = ((l[i] + x - 1) / x) * x;
    if (l1 > r1) {
      return;
    }
    if (i == 1) {
      for (j = l1; j <= r1; j += x) {
        f[j][i] = 1;
      }
    } else {
      for (j = x; j <= m; j += x) {
        cnt++;
        if (j <= l1) {
          continue;
        }
        if (j <= r1 + x) {
          f[j][i] = pre[j - l1][i - 1];
        } else {
          f[j][i] = su(pre[j - l1][i - 1], 998244353 - pre[j - r1 - x][i - 1]);
        }
      }
    }
    for (j = x; j <= m; j += x) {
      pre[j][i] = su(pre[j - x][i], f[j][i]);
    }
  }
  for (i = x; i <= m; i += x) {
    res = su(res, f[i][n]);
  }
  g[x] = res;
}
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++) {
    get(i);
  }
  for (i = m; i >= 1; i--) {
    for (j = i + i; j <= m; j += i) {
      g[i] = su(g[i], 998244353 - g[j]);
    }
  }
  printf("%d", g[1]);
}

```