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: Wrong Answer)

#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<map>
#include<string.h>
#include<complex>
#include<math.h>
#include<queue>
#include <functional>
#include<time.h>
#include <stack>
#include<iomanip>
using namespace std;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define ll long long
#define llint long long int
#define reverse(v)  reverse(v.begin(), v.end());
#define Yes(ans) if(ans)cout<<"Yes"<<endl; else cout<<"No"<<endl;
#define YES(ans) if(ans)cout<<"YES"<<endl; else cout<<"NO"<<endl;
#define hei(a) vector<a>
#define whei(a) vector<vector<a>>
#define UF  UnionFind 
#define Pint pair<llint,llint>
#define keta(a)  fixed << setprecision(a)
constexpr auto INF = 1000000000000000;
constexpr auto mod = 1000000007;
//辞書順はnext_permutation( begin( v ), end( v ) );やで！
// lower_boundは二分探索
struct edge { int to, cost; };
long long modpow(long long a, long long n, long long mod) {
	long long res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

// mod. m での a の逆元 a^{-1} を計算するよ！
long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}


//aCbをmod.mで割った余りを求める
ll int c(ll int a, ll int b, ll int m) {
	ll int ans = 1;
	for (ll int i = 0; i < b; i++) {
		ans *= a - i;
		ans %= m;
	}
	for (ll int i = 1; i <= b; i++) {
		ans *= modinv(i, m);
		ans %= m;
	}
	return ans;
}

//sは初めの点、vは頂点size、dは最短距離を求める(初期化はINF),Gは有向グラフ
void dijkdtra(int s, int v, vector<int>& d, whei(edge)& G) {
	priority_queue<Pint, vector<Pint>, greater<Pint>> que;//firstが距離、secondが頂点
	d[s] = 0;//スタートの距離はもちろんzero
	que.push(Pint(0, s));
	while (!que.empty()) {
		Pint p = que.top();
		que.pop();
		int V = p.second;
		if (d[V] < p.first)continue;
		for (int i = 0; i < G[V].size(); i++) {
			edge e = G[V][i];
			if (d[e.to] > d[V] + e.cost) {
				d[e.to] = d[V] + e.cost;
				que.push(Pint(d[e.to], e.to));
			}
		}
	}
}

llint binary_search(vector<int>&s,llint a) {
	llint l = -1;
	llint r = (int)s.size();
	while (r - l > 1) {
		llint mid = l + (r - l) / 2;

		if (s[mid]>=a) r = mid;
		else l = mid;
	}
	return r;
}
int k(ll n) {
	int x=0;
	while (n) {
		x+=n % 10;
		n /= 10;
	}
	return x;
}
ll max(ll x, ll y) {
	if (x < y)return y;
	return x;
}


int F(int n) {
	int x = 0;
	while (n) {
		x += n % 10;
		n /= 10;
	}
	return x;
}


int main() {
	llint n, G;
	int ans = (int)INF;
	cin >> n >> G;
	hei(int) p(n), c(n);
	rep(i, 0, n) {
		cin >> p[i] >> c[i];
	}
	rep(bit, 0, (1 << n)) {
		int sum = 0;
		int cost = 0;
		rep(i, 0, n) {
			if (bit & (1 << i)) {
				cost += p[i];
				sum += p[i] * 100 * (1+i);
			}
		}
		if (sum >= G)ans = min(ans, cost);
	}
	cout << ans << endl;
	return 0;
}