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>
#define maxD 12
using namespace std;
int p[maxD], c[maxD];
bool v[maxD];
int main(){
	int D, G;
	scanf("%d %d", &D, &G);
	for(int i = 1; i <= D; i += 1){
		scanf("%d %d", p + i, c + i);
		c[i] = p[i] * i  + c[i] / 100; 
	}
	int ans = 1000;
	for(int i = 0; i < (1 << D); i += 1){
		int pG = G / 100, pans = 0, k = 0;
		fill(v + 1, v + D + 1, false);
		for(int j = 1; j <= D; j += 1) if(i & (1 << j - 1)){
			v[j] = true;
			pG -= c[j];
			pans += p[j];
		}
		else k = j;
		for(int j = 1; j <= p[k] - 1 && pG > 0; j += 1){
			pG -= k;
			pans += 1;
		}
		if(pG <= 0) ans = min(ans, pans);
	}
	printf("%d", ans);
}