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 <vector>
#include <algorithm>
#include <string.h>
#include <iomanip>
#include <math.h>
#include <string>
#include <queue>
#include <map>
#include <set>
#include <functional>

using namespace std;

typedef long long int ll;
typedef std::pair<int, int> pii;
typedef std::pair<ll, int> pli;
typedef std::pair<ll, ll> pll;

#define FOR(i,n,m) for(ll i=(ll)(m);i<(ll)(n);++i)
#define REP(i,n) FOR(i,n,0)
#define IREP(i,n) for(ll i=(ll)(n);i>=0;--i)
#define OF64 std::setprecision(10)

const ll MOD = 1000000007;
const ll INF = (ll)1e15;

pll pc[15];

vector<vector<int>> s;
int d = 0;
int D;

void dfs(vector<int> v)
{
	if (v.size() == D)
	{
		s.push_back(v);
		return;
	}
	IREP(i, D - 1)
	{
		bool c = false;
		REP(j, v.size())
		{
			if (v[j] == i)
			{
				c = true;
				break;
			}
		}
		if (c)continue;

		v.push_back(i);
		dfs(v);
		v.pop_back();
	}
	return;
}

int main()
{

	ll G;
	cin >> D >> G;
	vector<ll> v(D);//! 全部解いたら何点？
	ll t = 1;
	REP(i, D)
	{
		t *= (i + 1);
		cin >> pc[i].first >> pc[i].second;
		v[i] = (i + 1) * pc[i].first * 100 + pc[i].second;
	}

	//s.resize(t);
	vector<int> tmp;
	dfs(tmp);


	ll ans = INF;

	REP(i, s.size())
	{
		ll sum = 0;
		ll sumI = 0;
		REP(j, D)
		{
			int index = s[i][j];
			ll n = (G - sum + 100 * (1 + index) - 1) / (100 * (1 + index));
			if (n < pc[index].first)
			{
				sum += 100 * (1 + index)*n;
				sumI += n;
			}
			else
			{
				sum += v[index];
				sumI += pc[index].first;
			}
			if (sum >= G)
			{
				ans = std::min(ans, sumI);
				break;
			}
			if (ans <= sumI)
			{
				break;
			}
		}
	}

	cout << ans << endl;

	return 0;
}