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.Scanner;
public class Main {

	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int n = in.nextInt(), goal = in.nextInt();
		Problem[] p = new Problem[n];
		for (int i=0; i<n; i++) {
			p[i] = new Problem(in.nextLong(), (i+1)*100, in.nextLong());
		}

		//ビット全探索+@
		long min = Long.MAX_VALUE;
		for (int i=0; i<(1<<n); i++) {

			int num = 0; //その組み合わせでは何問解いたか？
			long sum = 0L; //その組み合わせで得られたスコアの合計

			//全完していない問題のうち、最もスコアが高いもの、すなわち
			//i をビット表現したときに最も右側にある 0 の位置を記録する変数
			int rightMost = -1;

			for (int j=n-1; j>=0; j--) { //
				if ((1&i>>j) == 1) {
					sum += p[j].num * p[j].score + p[j].bonus;
					num += p[j].num;
				}
				else if ((1&i>>j) == 0 && rightMost == -1) {
					rightMost = j;
				}
			}

			//sum が目標スコア以上の場合は、解いた問題の数をそのまま min に記録する
			if (sum >= goal) {
				min = Math.min(min, num);
				continue;
			}
			else if (rightMost != -1){
				//そうでない場合かつ、全完していない問題が少なくとも1つある場合

				//目標スコアに届くまで、スコアが最も多い問題を needNum 回解く必要がある
				long needNum = (goal-sum+p[rightMost].score-1) / p[rightMost].score;

				if (needNum > p[rightMost].num) {
					//スコアが最も多い問題が needNum 問より少ないなら、放棄
					continue;
				}
				else {
					//そうでないなら、スコアが最も多い問題を needNum 問解き、
					//min を更新する
					min = Math.min(min, num + needNum);
				}
			}

		}

		System.out.println(min);

	}

	static class Problem {
		long num, score, bonus;
		Problem (long a, long b, long c) {
			num = a; score = b; bonus = c;
		}
	}


}