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 <iostream>
#include <vector>
#include <map>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <numeric>
#include <tuple>
#include <stack>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(x) x.begin(),x.end()
#define rall(x) x.rbegin(),x.rend()
#define print(s) cout << s << endl
using namespace std;
const int INF = 1e9;
typedef long long lint;
const lint LINF = 1e18;
const lint MOD = 1e9 + 7;
// const lint MOD = 998244353;
int ctoi(char c) {
	if (c >= '0' && c <= '9') {
		return c - '0';
	}
	return 0;
}
char upper(char c){
	return c-0x20;
}
char lower(char c){
	return c+0x20;
}

int gcd(int x, int y) {
  if (y == 0) return x;
  return gcd(y, x%y);
}
 
int main(){
	// D: 問題の種類
	// G: 要求得点
	// p: 各問題の個数
	// c: コンプリート得点
    int D, G, p[11], c[11];
    cin >> D >> G;
    for(int i = 0; i < D; ++i){
        cin >> p[i] >> c[i];
    }

    int ans = 1e9;
    for(int mask = 0; mask < (1 << D); ++mask){
		// s: p全て解いた時の得点
		// num: 解いた個数
		//reset_max: 解いてない問題における最大得点問題
        int s = 0, num = 0, rest_max = -1;
        for(int i = 0; i < D; ++i){
            if(mask >> i & 1){
                s += 100 * (i+1) * p[i] + c[i];
                num += p[i];
            }else{
                rest_max = i;
            }
        }
		//求められる得点以下ならば
        if(s < G){
			// s1: 
            int s1 = 100 * (rest_max + 1);
            int need = (G - s + s1 - 1) / s1;
            if(need >= p[rest_max]){
                continue;
            }
            num += need;
        }
        ans = min(ans, num);
    }
    cout << ans << endl;
}