Problem:
Score : 800 points

Problem Statement

There are N stores called Store 1, Store 2, \cdots, Store N. Takahashi, who is at his house at time 0, is planning to visit some of these stores.

It takes Takahashi one unit of time to travel from his house to one of the stores, or between any two stores.

If Takahashi reaches Store i at time t, he can do shopping there after standing in a queue for a_i \times t + b_i units of time. (We assume that it takes no time other than waiting.)

All the stores close at time T + 0.5. If Takahashi is standing in a queue for some store then, he cannot do shopping there.

Takahashi does not do shopping more than once in the same store.

Find the maximum number of times he can do shopping before time T + 0.5.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

0 \leq a_i \leq 10^9

0 \leq b_i \leq 10^9

0 \leq T \leq 10^9

Input

Input is given from Standard Input in the following format:

N T
a_1 b_1
a_2 b_2
\vdots
a_N b_N

Output

Print the answer.

Sample Input 1

3 7
2 0
3 2
0 3

Sample Output 1

2

Here is one possible way to visit stores:

From time 0 to time 1: in 1 unit of time, he travels from his house to Store 1.

From time 1 to time 3: for 2 units of time, he stands in a queue for Store 1 to do shopping.

From time 3 to time 4: in 1 unit of time, he travels from Store 1 to Store 3.

From time 4 to time 7: for 3 units of time, he stands in a queue for Store 3 to do shopping.

In this way, he can do shopping twice before time 7.5.

Sample Input 2

1 3
0 3

Sample Output 2

0

Sample Input 3

5 21600
2 14
3 22
1 3
1 10
1 9

Sample Output 3

5

Sample Input 4

7 57
0 25
3 10
2 4
5 15
3 22
2 14
1 15

Sample Output 4

3

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

#![allow(dead_code)]
#![allow(unused_macros)]

// 入力マクロ
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

// #[allow(unused_imports)]
// use std::time::Instant;
// macro_rules! measure {
//     ( $x:expr) => {{
//         let start = Instant::now();
//         let result = $x;
//         let end = start.elapsed();
//         println!("Running in : {}.{:03}s", end.as_secs(), end.subsec_millis());
//         result
//     }};
// }

// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
const MOD: u64 = 1_000_000_007;
fn main() {
    input! {
        n: usize, t: usize,
        wait: [(usize, usize); n]
    }
    let mut wait = wait;
    wait.sort_by_key(|i| i.1);

    let mut time = 1;
    let mut ans = 0;
    while time < t && ans < n {
        time += wait[ans].0 * time + wait[ans].1 + 1;
        ans += 1;
    }
    println!("{}", ans);
}


// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

/// プログラムを終了
fn exit() {
    use std::process;
    process::exit(0)
}

/// 比較可能な2つの引数のうち大きい方を返す
fn max<T: PartialOrd>(n: T, m: T) -> T {
    if n >= m {
        n
    } else {
        m
    }
}

/// 比較[可能な2つの引数のうち小さい方を返す
fn min<T: PartialOrd>(n: T, m: T) -> T {
    if n <= m {
        n
    } else {
        m
    }
}

/// floor + as i64
fn into_floor(a: f64) -> i64 {
    (a - a % 1.0) as i64
}

/// ceil + as i64
fn into_ceil(a: f64) -> i64 {
    (a - a % 1.0) as i64 + 1i64
}

/// "Yes", "No" を出力
fn yes_no(flag: bool) {
    if flag {
        println!("Yes");
    } else {
        println!("No");
    }
}

/// 互除法を用いて最大公約数を求める
fn gcd(n: u64, m: u64) -> u64 {
    let (mut n, mut m) = (max(n, m), min(n, m));
    if m == 0 {
        return n;
    }
    let mut _r = 0;
    while n % m != 0 {
        _r = n % m;
        n = m;
        m = _r;
    }
    m
}

/// 最小公倍数を求める
fn lcm(n: u64, m: u64) -> u64 {
    n * m / gcd(n, m)
}

/// 配列全体の最大公約数
fn gcd_vec(v: &[u64]) -> u64 {
    let mut r = v[0];
    for i in v.iter().skip(1) {
        r = gcd(r, *i);
    }
    r
}

/// 配列全体の最小公倍数
fn lcm_vec(v: &[u64]) -> u64 {
    let mut r = v[0];
    for i in v.iter().skip(1) {
        r = lcm(r, *i);
    }
    r
}

/// Find the first factor (other than 1) of a number
fn firstfac(x: u64) -> u64 {
    if x % 2 == 0 {
        return 2;
    };
    // TODO: return to step_by
    // for n in (3..).step_by(2).take_while(|m| m*m <= x) {
    for n in (1..).map(|m| 2 * m + 1).take_while(|m| m * m <= x) {
        if x % n == 0 {
            return n;
        };
    }
    // No factor found. It must be prime.
    x
}

/// Test whether a number is prime. Checks every odd number up to `sqrt(n)`.
fn is_prime(n: u64) -> bool {
    if n <= 1 {
        return false;
    }
    firstfac(n) == n
}

// TODO!
// pub fn parmutation<T: PartialOrd>(v: &Vec<T>) -> Vec<Vec<T>> {
//     let mut sorted = v;

// }