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>

#define rep2(x,fr,to) for(int (x)=(fr);(x)<(to);(x)++)
#define rep(x,to) for(int (x)=0;(x)<(to);(x)++)
#define repr(x,fr,to) for(int (x)=(fr);(x)>=(to);(x)--)
#define all(c) (c).begin(),(c).end()
#define sz(v) (int)(v).size()

using namespace std;
typedef long long ll; typedef vector<int> VI; typedef pair<int,int> pii;
const int mod = (int)1e9 + 7;
void dbg(){ cerr << "\n"; } template <typename T,typename ...T2> void dbg(const T& fst, const T2&...rst){ cerr << fst << ": "; dbg(rst...); }



int main()
{
	
	cin.tie(0); ios_base::sync_with_stdio(false);
	int d, g;
	cin >> d >>g;
	VI p(d), c(d);
	rep(i, d) cin >>p[i] >>c[i];
	int ans = 1<<29;
	rep(i, 1<<d){
		int pt =0, cx =0;
		rep(j, d) if(i>>j &1){
			pt +=c[j];
			pt += (j+1)* p[j] * 100;
			cx += p[j];
		}
		
		if(pt < g){
			repr(j, d-1, 0)if(!(i>>j &1)){
				if(g > pt + (j+1)*100*p[j]){
					pt += (j+1)*100*p[j]; cx += p[j];
				}else{
					int z = (g-pt + (j+1)*100 -1) / ((j+1)*100);
					pt += (j+1)*100*z; cx += z;
				}
				if(pt >= g) break;
			}
		}
		ans = min(ans, cx);
		
	}
	cout << ans <<"\n";
	return 0;
}