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: Time Limit Exceeded)

#include<iostream>
#include<cstdlib>
#include<cstring>
#include<stack>
#include<queue>
#include<algorithm>
#include<vector>
#include<math.h>
#include<bitset>
#include<iomanip>
#include<numeric>
using namespace std;

typedef stack<int> IntStack;
typedef queue<int> IntQueue;

const double pi = 3.141592653589793;


//最大公約数
template<class T> T gcd(T a, T b)
{
	while (true)
	{
		T r;
		if (a%b == 0) return b;
		else
		{
			r = a % b;
			a = b;
			b = r;
		}
	}
}
//最小公倍数
template<class T> T lcm(T a, T b)
{
	return a / gcd(a, b)*b;
}
//文字0~9をint型0~9に変換
int char_to_int(char c)
{
	if (c == '0') return 0;
	if (c == '1') return 1;
	if (c == '2') return 2;
	if (c == '3') return 3;
	if (c == '4') return 4;
	if (c == '5') return 5;
	if (c == '6') return 6;
	if (c == '7') return 7;
	if (c == '8') return 8;
	if (c == '9') return 9;
	else return -1;
}
//a^b mod MODを計算
long long int ModPower(long long int a, long long int b, long long int MOD)
{
	if (b == 0) return 1;
	else if (b % 2 == 0)
	{
		long long int d = ModPower(a, b / 2, MOD);
		return (d*d) % MOD;
	}
	else
	{
		return (a*ModPower(a, b - 1, MOD)) % MOD;
	}
}
//nCr mod MODを計算
long long int ModCombi(long long int n, long long int r, long long int MOD)
{
	if (n == 0 || r == 0) return 1;

	long long int Kn = 1, Kr = 1, Knr = 1;

	long long int temp = 1;
	for (long long int i = 1; i <= n; i++)
	{
		temp *= i;
		temp %= MOD;
		if (i == n) Kn = temp;
		if (i == r) Kr = temp;
		if (i == n - r) Knr = temp;
	}

	long long int x = ModPower(Kr, MOD - 2, MOD);
	long long int y = ModPower(Knr, MOD - 2, MOD);

	long long int ans;
	ans = Kn * x;
	ans %= MOD;
	ans *= y;
	ans %= MOD;

	return ans;
}


/*ここまでテンプレ*//*ここまでテンプレ*//*ここまでテンプレ*//*ここまでテンプレ*/

int D, G;
int p[15], c[15];
//必要な問題数を返す
int dfs(int sum, int d)
{
	if (sum <= 0) return 0;

	if (d + 1 == D)
	{
		if (sum > 100 * (d + 1) * p[d] + c[d]) return 2000;
		if (sum <= 100 * (d + 1) * p[d])
		{
			int a = sum / (100 * (d + 1));
			if (sum % (100 * (d + 1)) != 0) a++;
			return a;
		}
		return p[d];
	}

	int m = 2000;
	for (int i = 0; i < p[d]; i++)
	{
		m = min(m, i + dfs(sum - i * 100 * (d + 1), d + 1));
	}
	m = min(m, p[d] + dfs(sum - (d + 1) * 100 * p[d] - c[d], d + 1));

	return m;
}

int main()
{
	cin >> D >> G;
	for (int i = 0; i < D; i++)
	{
		cin >> p[i] >> c[i];
	}

	cout << dfs(G, 0) << endl;

	return 0;
}