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 <algorithm>
#include <cmath>
#include <iostream>
#include <numeric>
#include <string>
#include <string.h>
#include <vector>

#define REP(i,x) for(int i{ 0 }; i < (int)(x); i++)
#define REPC(i,x) for(int i{ 0 }; i <= (int)(x); i++)
#define RREP(i,x) for(int i{ (int)(x) - 1 }; i >= 0 ;i--)
#define RREPC(i,x) for(int i{ (int)(x)}; i >= 0; i--)
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(),(x).end()

using namespace std;

typedef int64_t int64;

const int dx[4] = { 1, 0, -1,  0 };
const int dy[4] = { 0, 1,  0, -1 };

template<class T>bool chmax(T& a, const T& b) { if (a < b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T& a, const T& b) { if (b < a) { a = b; return 1; } return 0; }

class C {
	int d, target;
	int mask{ 0 };
	vector<vector<int>> problems;
public:
	C()
	{
		cin >> d >> target;
		problems.resize(d, vector<int>(2));
		
		REP(i, d)
		{
			mask |= (1 << i);
			cin >> problems[i][0] >> problems[i][1];
			problems[i][1] += 100 * (i + 1) * problems[i][0];
		}
	}

	int calcNQuestions(int bit)
	{
		int ret{ 0 };
		for (int b{ 0 }; b < d; b++)
			if (bit & b)
				ret += problems[b][0];

		return ret;
	}

	void solve()
	{

		int ans{ 2000 };
		int sum;
		int n_questions;

		for (int bit{ 0 }; bit < (1 << d); bit++)
		{
			sum = 0;
			n_questions = 0;

			for (int b{ 0 }; b < d; b++)
				if (bit & (1 << b))
				{
					sum += problems[b][1];
					n_questions += problems[b][0];
				}

			if (sum >= target)
			{
				chmin(ans, n_questions);
				continue;
			}
			else
			{
				int notUsed{ mask ^ bit };
				if (notUsed)
				{
					// look for highest bit that is zero
					for (int b{ d - 1 }; b >= 0; b--)
					{
						if (notUsed & (1 << b))
						{
							// find out how many questions we need to the target
							int needMore{ ceil(((double)target - sum) / 100 / (b + 1)) };

							// if the number does not exceed existing qustions
							if (needMore <= problems[b][0])
								// and add the number to n_questions
								// update ans
								chmin(ans, n_questions + needMore);

							break;
						}
					}
				}
			}
		}

		cout << ans << endl;
	}
};


int main()
{
	C solution;
	solution.solve();

	return 0;
}