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 <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cassert>
#include <climits>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <fstream>

using namespace std;
#define TOSTRING(x) #x
#define SZ(x) (int)(x).size()
#define REP(i,n) for(int i=0;i<(n);i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REPR(i,n) for(int i=(n)-1;i>=0;i--)
#define ALL(s) (s).begin(), (s).end()
#define so(V) sort(ALL(V))
#define rev(V) reverse(ALL(V))
#define uni(v) v.erase( unique(ALL(v)) , v.end());
#define PAU system("pause")

typedef long long unsigned int llu;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vector <int> > vvi;
const int EPS = 1e-9;
const int MOD = 1e9 + 7;
const int INF = (1 << 30);
const double PI = acos(-1);

typedef pair<int, int> P;

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int D, G;
	cin >> D >> G;
	vi P(D), C(D);
	REP(i, D)cin >> P[i] >> C[i];
	int ans = INT_MAX;
	REP(i, (1 << D)) {//ビットマスク
		int score = 0;
		int cnt = 0;
		REP(j, D) {
			if ((i & (1 << j)) != 0) {
				score += (j + 1) * 100 * P[j] + C[j];
				cnt += P[j];
			}
		}
		if (score >= G) {
			ans = min(ans, cnt);
			continue;
		}
		REPR(j, D) {
			int ii = i & (1 << j);
			if ((i & (1 << j)) == 0) {
				int ind = 0;
				while (score < G && ind < P[j]) {
					score += (j + 1) * 100;
					cnt++; ind++;
				}
			}
		}
		if (score >= G) {
			ans = min(ans, cnt);
		}
	}
	cout << ans << endl;
	PAU;
	return 0;
}