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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;

/*Atcoder用
 * 		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] tmp = br.readLine().split(" ");
		int a = Integer.parseInt(tmp[0]);
		int b = Integer.parseInt(tmp[1]);
		int c = Integer.parseInt(tmp[2]);


		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String get = br.readLine();
		char[] s = get.toCharArray();
		for(char c:s)System.out.println(c);
 */

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] tmp = br.readLine().split(" ");
		int D = Integer.parseInt(tmp[0]);
		int G = Integer.parseInt(tmp[1]);

		ArrayList<Integer> p = new ArrayList<Integer>();
		ArrayList<Integer> c = new ArrayList<Integer>();
		int max_dec = (int) Math.pow(2, D) - 1;

		for (int d = 0; d < D; d++) {
			tmp = br.readLine().split(" ");
			int pp = Integer.parseInt(tmp[0]);
			int cc = Integer.parseInt(tmp[1]);
			p.add(pp);
			c.add(cc);
		}
		int best = 99999999;

		// TODO:D=1の処理
		// 求めて出力、exit(0);

		if(D==1){
			int point=0;
			int count=0;
			for(int i = 0;i<p.get(0)-1;i++){
				point += 100;
				count++;
				if(point>=G){
					System.out.println(count);
					System.exit(0);
				}
			}
			System.out.println(p.get(0));
			System.exit(0);
		}

		for (int i = 1; i < max_dec; i++) {
			String bin = Integer.toBinaryString(i);

			while (bin.length() < D) {
				bin = "0" + bin;
			}
			int point = 0;
			int solvedProblem = 0;
			int maxZero = 0; // 0を持つ最大桁を探索する


			for (int j = 0; j < bin.length(); j++) {
				int b = bin.charAt(bin.length() - j - 1);
				if (b == 49) {
					boolean isFull = false;
					for (int r = 0; r < p.get(j); r++) {
						if (point >= G) {
							isFull = true;
							break;
						}
						point += (j + 1) * 100;
						solvedProblem++;
					}

					if (!isFull) {
						point += c.get(j);
					}
				} else {
					maxZero = j;
				}
			}

			if (point < G) {
				for (int j = 0; j < p.get(maxZero) - 1; j++) {
					point += (maxZero + 1) * 100;
					solvedProblem++;
					if (point >= G)
						break;
				}

				if (point < G)
					solvedProblem = 99999999;
			}

			best = Math.min(solvedProblem, best);

		}
		System.out.println(best);

	}

}