Problem:
Score : 300 points

Problem Statement

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).

Sample Input 1

2 700
3 500
5 800

Sample Output 1

3

In this case, there are three problems each with 100 points and five problems each with 200 points. The perfect bonus for solving all the 100-point problems is 500 points, and the perfect bonus for solving all the 200-point problems is 800 points. Takahashi's objective is to have a total score of 700 points or more.

One way to achieve this objective is to solve four 200-point problems and earn a base score of 800 points. However, if we solve three 100-point problems, we can earn the perfect bonus of 500 points in addition to the base score of 300 points, for a total score of 800 points, and we can achieve the objective with fewer problems.

Sample Input 2

2 2000
3 500
5 800

Sample Output 2

7

This case is similar to Sample Input 1, but the Takahashi's objective this time is 2000 points or more. In this case, we inevitably need to solve all five 200-point problems, and by solving two 100-point problems additionally we have the total score of 2000 points.

Sample Input 3

2 400
3 500
5 800

Sample Output 3

2

This case is again similar to Sample Input 1, but the Takahashi's objective this time is 400 points or more. In this case, we only need to solve two 200-point problems to achieve the objective.

Sample Input 4

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

Sample Output 4

66

There is only one 500-point problem, but the perfect bonus can be earned even in such a case.

Solution in C++14 (GCC 5.4.1) (judge verdict: Memory Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define ing long long
#define INF 184497440
#define MMM 20000000

int temp[20][MMM] = {};

signed main(){
	int D, G, te[2][20] = {}, MA=0, ans = INF;
	cin>>D>>G; G/=100;
	for(int i = 0; i < 20; i++)
		for(int j = 1; j < MMM; j++)temp[i][j] = INF;
	temp[0][0] = 0;
	for(int i = 1; i <= D; i++){
		cin>>te[0][i]>>te[1][i];
		te[1][i] /= 100;
		MA += te[0][i]*i+te[1][i];
	}
	
	for(int i = 1; i <= D; i++){//cout<<i<<G<<"---"<<endl;
		for(int j = 0; j <= G+1; j++){//cout<<temp[i][j]<<"  ";
			
			if(temp[i-1][j] == INF)continue;//cout<<i<<" "<<j<<endl;
			for(int k = 1; k <= te[0][i]; k++){//cout<<"k = "<<k<<endl;
				if(k==te[0][i]){
					//if(j < te[0][i])
					//cout<<" AA "<<j<<"+"<<te[1][i]<<"+"<<i*te[0][i]<<" "<<temp[i-1][j] + te[0][i]<<endl;
					temp[i][j+te[1][i] + i*te[0][i]] = min(temp[i][j+te[1][i] + i*te[0][i]], temp[i-1][j] + te[0][i]);
				}else{
					//if(j < i)continue;
					//cout<<temp[i-1][j]<<" A "<<k<<endl;
					temp[i][j+k*i] = min(temp[i][j+k*i], temp[i-1][j]+k);
				}
			}
			temp[i][j] = min(temp[i][j], temp[i-1][j]);
		}//cout<<endl;
	}//cout<<endl;
	/*for(int j = 1; j <= D; j++){
	for(int i = 0; i <= 10; i++){
		cout<<i<<" "<<temp[j][i]<<endl;
	}
	}
	*/
	for(int j = G; j < MMM; j++){
		ans = min(ans, temp[D][j]);
	}
	cout<<ans<<endl;
	return 0;
}