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).

Examples

Input

2 700
3 500
5 800


Output

3


Input

2 2000
3 500
5 800


Output

7


Input

2 400
3 500
5 800


Output

2


Input

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


Output

66
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

int D, G;
int f[20][1010];

int p[20], c[20];

int main(){
	cin >> D >> G;
	
	for(register int i=1; i<=D; i++){
		cin>>p[i]>>c[i];
	}
	
	for(register int i=1; i<=D; i++){
		for(register int j=0; j<p[i]; j++){
			for(register int k=j; k<=1000; k++)
				f[i][k] = max(f[i][k], f[i-1][k-j] + j *(100 * i));
		}
		
		for(register int k=p[i]; k<=1000; k++)
			f[i][k] = max(f[i][k], f[i-1][k-p[i]] + p[i] * (100 * i) + c[i]);
	}
	
	int ans = 1000;
	for(register int i=1000; i>=0; i--)
		if(f[D][i] >= G)
			ans = i;
	
	printf("%d", ans);
}