Mocha and Stars solution codeforces

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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 $(a_{1}, a_{2}, \dots, a_{n})$ to show its state. A state is called mathematical if all of the following three conditions are satisfied:

For all $i$ ( $1 \leq i \leq n$ ), $a_{i}$ is an integer in the range $[l_{i}, r_{i}]$ .
$\sum_{i = 1}^{n} a_{i} \leq m$ .
$gcd (a_{1}, a_{2}, \dots, a_{n}) = 1$ .

Here, $gcd (a_{1}, a_{2}, \dots, a_{n})$ denotes the greatest common divisor (GCD) of integers $a_{1}, a_{2}, \dots, a_{n}$ .

Mocha is wondering how many different mathematical states of this constellation exist. Because the answer may be large, you must find it modulo $998 244 353$ . Mocha and Stars solution codeforces

Two states $(a_{1}, a_{2}, \dots, a_{n})$ and $(b_{1}, b_{2}, \dots, b_{n})$ are considered different if there exists $i$ ( $1 \leq i \leq n$ ) such that $a_{i} \neq b_{i}$ .

Input

The first line contains two integers $n$ and $m$ ( $2 \leq n \leq 50$ , $1 \leq m \leq 10^{5}$ ) — 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 $r_{i}$ ( $1 \leq l_{i} \leq r_{i} \leq m$ ) — 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 Mocha and Stars solution codeforces

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2 4
1 3
1 2

output

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input

Copy

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Copy Mocha and Stars solution codeforces

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output

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47464146

Note

In the first example, there are $4$ different mathematical states of this constellation:

$a_{1} = 1$ , $a_{2} = 1$ .
$a_{1} = 1$ , $a_{2} = 2$ .
$a_{1} = 2$ , $a_{2} = 1$ .
$a_{1} = 3$ , $a_{2} = 1$ .
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Mocha and Stars solution codeforces

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