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).

Examples

Input

2 700
3 500
5 800


Output

3


Input

2 2000
3 500
5 800


Output

7


Input

2 400
3 500
5 800


Output

2


Input

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000


Output

66
import java.util.*;
import java.util.function.*;
import java.math.*;
public class Main {
    public static void main(String[] args){
		Scanner s = new Scanner(System.in);
		int D = s.nextInt();
		int G = s.nextInt();
		int[] p = new int[D]; 
		int[] c = new int[D]; 
		for(int i = 0; i < D; i++) {
			p[i] = s.nextInt();
			c[i] = s.nextInt();
		}
		
		boolean[] complete = new boolean[D];
		int[][] score = new int[D][];
		for(int i = 0; i < D; i++) {
			score[i] = new int[p[i]];
			for(int j = 0; j < p[i]; j++) {
				score[i][j] = (100 * (i + 1)) * (j + 1) + (j == p[i] - 1 ? c[i] : 0);
			}
		}
		
		// 引数1：必要点数 戻り値：必要点数を取得するのに必要な問題数
		final BiFunction<Integer, boolean[], Integer> up = (i, fcomplete) -> {
			int count = 0;
			// 総問題数分繰り返す（1 ～ D 問）
			for(int j = D - 1; j > -1; j--) {
				if(fcomplete[j])
					continue;
				// j + 1 問目の問題をk問解いたときの点数をマイナスする
				for(int k = 1; k < p[j]; k++) {
					i -= 100 * (j + 1);
					count++;
					if(i <= 0) {
						return count;
					}
				}
			}
			return -1;
		};
		
		// 初期設定
		int result = Integer.MAX_VALUE;
		int candidate = 0;
		int count = 0;
		for(int i = 0; i < (1<<D); i++, count = 0, candidate = 0, complete = new boolean[D]) {
			// j = j + 1 問目の問題を解いたかいなか
			for(int j = 0; j < D; j++) {
				if(((i>>j)&1) == 1) {
					candidate += score[j][score[j].length - 1];
					count += p[j];
					complete[j] = true;
				}
			}
			int additinal = 0;
			if(candidate < G) {
				additinal = up.apply(G - candidate, complete);
				if(additinal == -1)
					continue;
			}
			result = Math.min(result, count + additinal);
		}
		
		System.out.println(result);
    }
}