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<bits/stdc++.h>
using namespace std;
int n;
long long INF = 200000;
long long dp[20][200000];
int bonus[20];
int p[20];
long long f(int curr,int score){
	if(score<=0) return 0;
	long long&ret = dp[curr][score];
	if(~ret) return ret;
	if(curr>n) return ret = INF;
	ret = INF;
	for(int i=0; i<p[curr]; i++){
		ret = min(ret,i+f(curr+1,score-i*curr));
	}
	ret = min(ret,p[curr]+f(curr+1,score-curr*p[curr]-bonus[curr]));
	return ret;
}
int main(){
	memset(dp,-1,sizeof(dp));
	int g;
	cin>>n>>g;
	g/=100;
	for(int i=1; i<=n; i++){
		cin>>p[i]>>bonus[i];
		bonus[i]/=100;
	}
	cout<<f(1,g);
}