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: Accepted)

	#include<bits/stdc++.h>
	#include<iostream>

	using namespace std;

	#define forn(i,n) for ( int i=0 ; i < n ; i++ )
	#define all(a) (a).begin(), (a).end()
	#define lb lower_bound
	#define ub upper_bound
	#define vi vector<int >
	#define pii pair<int, int>
	#define F first
	#define S second
	#define MP make_pair
	#define PB push_back
	#define SQ(a) (a)*(a)
	#define ll long long
	#define lli long long int
	#define ull unsigned long long
	#define endl '\n'
	#define Pi(a) printf("%d\n",a);
	#define debug2(a,b) printf("%d\t%d\n",a,b);
	#define debug1(a) printf("%d\n",a);
	#define Si(a) scanf("%d",&a);
	#define mod 1000000007
	#define w(n) while(n--)

	int d,p[12],c[12],g;
	int recurse(int place,bool visited[],int sum){
		int ans=-1;
		if(place>d) return 0;
		for(int i=1;i<=d;i++){
			//cout << i << " " << visited[i] << " ";
			if(!visited[i]){
				visited[i]=true;
				if(p[i]*i+c[i]+sum>g){
					int temp = g-sum;
					int req=temp/i + temp%i;
					int ht = req>=p[i]?p[i]:req;
					if(ans==-1) ans=ht;
					else ans=min(ans,ht);
				}
				else if(p[i]*i+c[i]+sum==g){
					if(ans==-1) ans=p[i];
					else ans=min(ans,p[i]);
				}
				else{
					int here=p[i]+recurse(place+1,visited,p[i]*i+c[i]+sum);
					if(ans==-1) ans=here;
					else ans=min(ans,here);
				}
				visited[i]=false;
				//cout << i << " " << ans << endl;
			}
		}
		return ans;
	}

	int main()
	{
		//ios::sync_with_stdio(0);
		//cin.tie(0);
		//code
		cin >> d >> g;
		g/=100;
		forn(i,d) cin >> p[i+1] >> c[i+1];
		forn(i,d) c[i+1]/=100;
		bool visited[d+1];
		memset(visited,false,sizeof(visited));
		cout << recurse(1,visited,0) << endl;
		return 0;

	}