Problem:
Score : 300 points

Problem Statement

A programming competition site AtCode provides algorithmic problems.
Each problem is allocated a score based on its difficulty.
Currently, for each integer i between 1 and D (inclusive), there are p_i problems with a score of 100i points.
These p_1 + … + p_D problems are all of the problems available on AtCode.

A user of AtCode has a value called total score.
The total score of a user is the sum of the following two elements:

Base score: the sum of the scores of all problems solved by the user.

Perfect bonuses: when a user solves all problems with a score of 100i points, he/she earns the perfect bonus of c_i points, aside from the base score (1 ≤ i ≤ D).

Takahashi, who is the new user of AtCode, has not solved any problem.
His objective is to have a total score of G or more points.
At least how many problems does he need to solve for this objective?

Constraints

1 ≤ D ≤ 10

1 ≤ p_i ≤ 100

100 ≤ c_i ≤ 10^6

100 ≤ G

All values in input are integers.

c_i and G are all multiples of 100.

It is possible to have a total score of G or more points.

Input

Input is given from Standard Input in the following format:

D G
p_1 c_1
:
p_D c_D

Output

Print the minimum number of problems that needs to be solved in order to have a total score of G or more points. Note that this objective is always achievable (see Constraints).

Sample Input 1

2 700
3 500
5 800

Sample Output 1

3

In this case, there are three problems each with 100 points and five problems each with 200 points. The perfect bonus for solving all the 100-point problems is 500 points, and the perfect bonus for solving all the 200-point problems is 800 points. Takahashi's objective is to have a total score of 700 points or more.

One way to achieve this objective is to solve four 200-point problems and earn a base score of 800 points. However, if we solve three 100-point problems, we can earn the perfect bonus of 500 points in addition to the base score of 300 points, for a total score of 800 points, and we can achieve the objective with fewer problems.

Sample Input 2

2 2000
3 500
5 800

Sample Output 2

7

This case is similar to Sample Input 1, but the Takahashi's objective this time is 2000 points or more. In this case, we inevitably need to solve all five 200-point problems, and by solving two 100-point problems additionally we have the total score of 2000 points.

Sample Input 3

2 400
3 500
5 800

Sample Output 3

2

This case is again similar to Sample Input 1, but the Takahashi's objective this time is 400 points or more. In this case, we only need to solve two 200-point problems to achieve the objective.

Sample Input 4

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000

Sample Output 4

66

There is only one 500-point problem, but the perfect bonus can be earned even in such a case.

Solution in OCaml (4.02.3) (judge verdict: Wrong Answer)

open Printf
open Scanf

let (d,g) = sscanf (read_line ()) "%d %d" (fun d g -> (d,g));;
let rec input_list num l =
if num = 0 then l
else 
    let (p,c) = sscanf (read_line ()) "%d %d" (fun d g -> (d,g)) in
    input_list (num-1) ((100 * (d-num+1), p, c) :: l);;

let list = input_list d [];;
let inf = 10000000000;;

let solve list g =
    let rec loop l q_num sum tl =
    match l with
    | [] -> 
        (match tl with
        | [] ->  if sum >= g then q_num else inf
        | x::rest -> 
            let diff = g - sum in
            if diff <= 0 then q_num 
            else 
                let (i1,p1,c1) = List.hd (List.rev tl) in
                if i1*(p1-1)+sum < g then inf
                else q_num + truncate ((float_of_int diff) /. (float_of_int i1))
        )
    | (i1,p1,c1)::rest -> 
        let used_value = loop rest (q_num + p1) (sum + p1*i1 + c1) tl in
        let non_used_value = loop rest q_num sum ((i1,p1,c1) :: tl) in
        min used_value non_used_value
    in loop list 0 0 [];;

let ans = solve list g;;

let _ = printf "%d\n" ans;;