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;
const int N=1e6+7;
#define ll long long
int f[20][N];
int p[N],c[N];
int main(){
	int d,g;
	scanf("%d%d",&d,&g);
	g/=100;
	for(int i = 1;i<=d;i++){
		scanf("%d%d",&p[i],&c[i]);
		c[i]/=100;
	}
	memset(f,0x3f3f3f3f,sizeof(f));
	for(int i = 1;i<=d;i++) f[i][0]=0;
	f[0][0]=0;
	for(int i = 1;i<=d;i++){
		for(int j = 1;j<=g;j++){
			for(int k = 0;k<=p[i];k++){
				int tmp=max(0,j-k*i);
				{
					f[i][j]=min(f[i][j],f[i-1][tmp]+k);
				}
				if(k==p[i]){
					int tmp=max(0,j-k*i-c[i]);
					{
						f[i][j]=min(f[i][j],f[i-1][tmp]+k);
					}
				}
			}
		}
	}
	printf("%d\n",f[d][g]);
	
}