Problem:
Score: 500 points

Problem Statement

N programmers are going to participate in the preliminary stage of DDCC 20XX. Due to the size of the venue, however, at most 9 contestants can participate in the finals.

The preliminary stage consists of several rounds, which will take place as follows:

All the N contestants will participate in the first round.

When X contestants participate in some round, the number of contestants advancing to the next round will be decided as follows:

The organizer will choose two consecutive digits in the decimal notation of X, and replace them with the sum of these digits. The number resulted will be the number of contestants advancing to the next round.

For example, when X = 2378, the number of contestants advancing to the next round will be 578 (if 2 and 3 are chosen), 2108 (if 3 and 7 are chosen), or 2315 (if 7 and 8 are chosen).

When X = 100, the number of contestants advancing to the next round will be 10, no matter which two digits are chosen.

The preliminary stage ends when 9 or fewer contestants remain.

Ringo, the chief organizer, wants to hold as many rounds as possible.
Find the maximum possible number of rounds in the preliminary stage.

Since the number of contestants, N, can be enormous, it is given to you as two integer sequences d_1, \ldots, d_M and c_1, \ldots, c_M, which means the following: the decimal notation of N consists of c_1 + c_2 + \ldots + c_M digits, whose first c_1 digits are all d_1, the following c_2 digits are all d_2, \ldots, and the last c_M digits are all d_M.

Constraints

1 \leq M \leq 200000

0 \leq d_i \leq 9

d_1 \neq 0

d_i \neq d_{i+1}

c_i \geq 1

2 \leq c_1 + \ldots + c_M \leq 10^{15}

Input

Input is given from Standard Input in the following format:

M
d_1 c_1
d_2 c_2
:
d_M c_M

Output

Print the maximum possible number of rounds in the preliminary stage.

Sample Input 1

2
2 2
9 1

Sample Output 1

3

In this case, N = 229 contestants will participate in the first round. One possible progression of the preliminary stage is as follows:

229 contestants participate in Round 1, 49 contestants participate in Round 2, 13 contestants participate in Round 3, and 4 contestants advance to the finals.

Here, three rounds take place in the preliminary stage, which is the maximum possible number.

Sample Input 2

3
1 1
0 8
7 1

Sample Output 2

9

In this case, 1000000007 will participate in the first round.

Solution in Rust (1.15.1) (judge verdict: Accepted)

use std::io;
use io_ext::Reader;
use parse::ParseAll;


fn main() {
    let stdin = io::stdin();
    let mut r = Reader::new(stdin.lock());
    let m: usize = r.read_line().parse().unwrap();
    let mut digits: Vec<i64> = Vec::with_capacity(m);
    let mut counts: Vec<i64> = Vec::with_capacity(m);
    for _ in 0..m {
        let (d, c) = r.read_line().split_whitespace().parse_all();
        digits.push(d);
        counts.push(c);
    }
    // let digits = vec![1, 0, 7];
    // let counts = vec![1, 8, 1];

    let ans = answer(&digits, &counts, 0);
    println!("{}", ans);
}

fn answer(digits: &[i64], counts: &[i64], mut accum: i64) -> i64 {
    let mut sum = 0;
    for (&d, &c) in digits.iter().zip(counts.iter()) {
        sum += d * c;
        accum += c;
    }
    accum -= 1;

    let ones = sum / 10;
    let last = sum % 10;
    if ones == 0 {
        accum
    } else {
        let digits = vec![1, last];
        let counts = vec![ones, 1];
        answer(&digits, &counts, accum)
    }
}

/// A module for easy use of io.
pub mod io_ext {
    use std::io::BufRead;

    pub struct Reader<R> {
        buf: String,
        inner: R,
    }

    impl<R> Reader<R> {
        #[inline]
        pub fn new(inner: R) -> Self {
            Reader { buf: String::new(), inner: inner }
        }

        #[inline]
        pub fn into_inner(self) -> R {
            self.inner
        }
    }

    impl<R: BufRead> Reader<R> {
        #[allow(deprecated)]
        #[inline]
        pub fn read_line(&mut self) -> &str {
            self.buf.clear();
            self.inner.read_line(&mut self.buf).unwrap_or_else(|e| panic!("{}", e));
            self.buf.trim_right()
        }
    }
}

/// Parsing Iterator.
pub mod parse {
    use std::borrow::Borrow;
    use std::str::FromStr;

    pub trait FromStrIterator {
        fn from_str_iter<S: Borrow<str>, I: Iterator<Item=S>>(i: I) -> Self;
    }

    pub trait ParseAll {
        fn parse_all<F: FromStrIterator>(self) -> F;
    }

    impl<S: Borrow<str>, I: Iterator<Item=S>> ParseAll for I {
        #[inline]
        fn parse_all<F: FromStrIterator>(self) -> F {
            F::from_str_iter(self)
        }
    }

    fn parse<S: Borrow<str>, I: Iterator<Item=S>, F: FromStr>(i: &mut I) -> F {
        i.next().unwrap_or_else(|| panic!("too few strings error")).borrow().parse().unwrap_or_else(|_| panic!("parse error"))
    }

    // To avoid conflict, this is not implemented for `A` but `(A,)`.
    impl<A: FromStr> FromStrIterator for (A, ) {
        fn from_str_iter<S: Borrow<str>, I: Iterator<Item=S>>(mut i: I) -> Self {
            let a = parse(&mut i);
            if i.next().is_some() {
                panic!("too many strings error");
            }
            (a, )
        }
    }

    impl<A: FromStr, B: FromStr> FromStrIterator for (A, B) {
        fn from_str_iter<S: Borrow<str>, I: Iterator<Item=S>>(mut i: I) -> Self {
            let a = parse(&mut i);
            let b = parse(&mut i);
            if i.next().is_some() {
                panic!("too many strings error");
            }
            (a, b)
        }
    }

    impl<A: FromStr, B: FromStr, C: FromStr> FromStrIterator for (A, B, C) {
        fn from_str_iter<S: Borrow<str>, I: Iterator<Item=S>>(mut i: I) -> Self {
            let a = parse(&mut i);
            let b = parse(&mut i);
            let c = parse(&mut i);
            if i.next().is_some() {
                panic!("too many strings error");
            }
            (a, b, c)
        }
    }

    impl<T: FromStr> FromStrIterator for Vec<T> {
        fn from_str_iter<S: Borrow<str>, I: Iterator<Item=S>>(i: I) -> Self {
            i.map(|s| s.borrow().parse().unwrap_or_else(|_| panic!("parse error"))).collect()
        }
    }
}

pub mod modulo {
    /// Returns `x` + `y` mod `modulo`.
    ///
    /// `x < modulo` and `y < modulo` must hold.
    #[inline]
    pub fn add(x: u32, y: u32, modulo: u32) -> u32 {
        debug_assert!(modulo > 0);
        debug_assert!(x < modulo && y < modulo);
        let sum = x as u64 + y as u64;
        if sum <= modulo as u64 {
            sum as u32
        } else {
            (sum as u32).wrapping_sub(modulo)
        }
    }

    /// Returns `x` - `y` mod `modulo`.
    ///
    /// `x < modulo` and `y < modulo` must hold.
    #[inline]
    pub fn sub(x: u32, y: u32, modulo: u32) -> u32 {
        debug_assert!(0 < modulo);
        debug_assert!(x < modulo && y < modulo);
        if x >= y {
            x - y
        } else {
            modulo + x - y
        }
    }

    /// Returns `x` * `y` mod `modulo`.
    #[inline]
    pub fn mul(x: u32, y: u32, modulo: u32) -> u32 {
        ((x as u64 * y as u64) % modulo as u64) as u32
    }

    /// Returns `x`^ `y` mod `modulo`.
    pub fn pow(x: u32, mut y: u32, modulo: u32) -> u32 {
        debug_assert!(0 < modulo);
        let mut p = x;
        let mut ret = 1;

        while y != 0 {
            if y & 1 == 1 {
                ret = mul(ret, p, modulo);
            }
            p = mul(p, p, modulo);
            y >>= 1;
        }
        ret
    }

    fn is_prime(x: u32) -> bool {
        let sqrt = (x as f32).sqrt() as u32;
        for factor in 2..sqrt {
            if x % factor == 0 {
                return false;
            }
        }
        true
    }

    /// Returns 1 / `x`.
    ///
    /// `modulo` must be a prime number.
    #[inline]
    fn reciprocal(x: u32, modulo: u32) -> u32 {
        debug_assert!(0 < modulo && is_prime(modulo));
        pow(x, modulo - 2, modulo)
    }

    /// Returns `x` / `y` mod `modulo`.
    ///
    /// `modulo` must be a prime number.
    #[inline]
    pub fn div(x: u32, y: u32, modulo: u32) -> u32 {
        debug_assert!(0 < modulo && is_prime(modulo));
        mul(x, reciprocal(y, modulo), modulo)
    }

    /// Returns `n`P`k` mod `modulo`.
    pub fn perm(n: u32, k: u32, modulo: u32) -> u32 {
        let mut p = 1;
        for i in 0..k {
            p = mul(p, n - i, modulo);
        }
        p
    }

    /// Returns `n`C`k` mod `modulo`.
    pub fn comb(n: u32, k: u32, modulo: u32) -> u32 {
        let num = perm(n, k, modulo);
        let mut den = 1;
        for i in 0..k {
            den = mul(den, i + 1, modulo);
        }
        div(num, den, modulo)
    }
}