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 <cstring>
#include <cstdio>

#define R register
#define Maxn 100010

using namespace std;

typedef long long ll;
int n , g , N , sum;
int p[Maxn] , c[Maxn];
ll f[Maxn];

int main() {
	
	scanf("%d%d",&n,&g);
	for(R int i=1; i<=n; i++) {
		scanf("%d%d",&p[i],&c[i]);
		sum += p[i];
	}
	
	for(R int i=1; i<=n; i++) {
		for(R int j=sum; j>=0; j--) {
			for(R int k=1; k<=p[i]; k++) {
				if(j-k < 0)	break;
				if(k < p[i])	f[j] = max(f[j],f[j-k]+ll(100*i*k));
				else	f[j] = max(f[j],f[j-k]+ll(100*i*k)+(ll)c[i]);
			}
		}
	}
	
	for(R int i=1; i<=sum; i++)
		if(f[i] >= g)
			return printf("%d\n",i) , 0;
	
	return 0;
	
}