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 REP(i, s,  n) for(int i = s;i < (n);i++)
#define VSORT(v) sort(v.begin(), v.end());
#define int64 long long
#define pb(a) push_back(a)
#define INF 999999999
using namespace std;

typedef pair<int, int> P;
typedef pair<int, P> PP;

int dy[]={0, 0, 1, -1, 0};
int dx[]={1, -1, 0, 0, 0};

//---------------------------------------------------------------
// 
//---------------------------------------------------------------

int D;
int64 G;
int p[10], c[10];		// 各点数の問題数とボーナス点

int count(int i=0, int cnt=0, int64 total=0, int v=-1) {
//    cout << "i = " << i << "  cnt = " << cnt << "  total = " << total << "  v = " << v << endl;
	if(i >= D) {
		if(total >= G)	{
//		    cout << "ret 1 : " << cnt << endl;
		    return cnt;
		}
		if(v >= 0) {
			int n = (G - total + (100 * (v+1)) - 1) / (100 * (v+1));
			if(n > p[v]) {
//    		    cout << "ret 2 : " << "n = " << n << " > " << p[v] << endl;
			    return INF;
			} else {
//    		    cout << "ret 3 : " << (cnt + n) << endl;
			    return (cnt + n);
			}
		} else {
//		    cout << "ret 4 : " << INF << endl;
		    return INF;
		}
	}
	
	int ans1 = count(i+1, cnt, total, i);
    int ans2 = count(i+1, (cnt + p[i]), total + ((i+1) * 100 * p[i] + c[i]), v);
    return min(ans1, ans2);
}

signed main(){
	cin >> D >> G;
	REP(i, 0, D)	cin >> p[i] >> c[i];
	
	cout << count() << endl;

	return 0;
}