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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var
    d,g,i,top,k,j:longint;
    a,f:array[0..200010] of longint;
    p,c:array[0..20] of longint;
function min(a,b:longint):longint;
begin
    if a<b then exit(a) else exit(b);
end;
function max(a,b:longint):longint;
begin
    if a>b then exit(a) else exit(b);
end;
begin
    readln(d,g);g:=g div 100;
    top:=1;a[top]:=g;
    for i:=1 to d do
        begin
            readln(p[i],c[i]);
            c[i]:=c[i] div 100;
            k:=top;
            while k>0 do
                begin
                    for j:=1 to p[i]-1 do
                    if f[max(a[k]-i*j,0)]=0 then 
                        begin
                            inc(top);a[top]:=max(a[k]-i*j,0);
                            f[max(a[k]-i*j,0)]:=f[a[k]]+j;
                        end
                    else f[max(a[k]-i*j,0)]:=min(f[max(a[k]-i*j,0)],f[a[k]]+j);
                    if f[max(a[k]-i*p[i]-c[i],0)]=0 then 
                        begin
                            inc(top);a[top]:=max(a[k]-i*p[i]-c[i],0);
                            f[max(a[k]-i*p[i]-c[i],0)]:=f[a[k]]+p[i];
                        end
                    else f[max(a[k]-i*p[i]-c[i],0)]:=min(f[max(a[k]-i*p[i]-c[i],0)],f[a[k]]+p[i]);
                    dec(k);
                end;
        end;
    for i:=1 to 10 do writeln(f[i]);
    writeln(f[0]);
end.