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)

/*
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*/
#include "bits/stdc++.h"
using namespace std;
#define ok1 printf("ok1\n");
#define ok2 printf("ok2\n");
#define M 1000000000000000000LL
#define rep(i,n) for(int i=0;i<n;++i)
#define REP(i,s,n) for(int i=(s);i<(n);++i)
#define repr(i,n) for(int i=n-1;i>=0;--i)
#define REPR(i,s,n) for(int i=(s);i>=(n);--(i))
#define all(a) (a).begin(),(a).end()
#define reall(a) (a).rbegin(),(a).rend()
#define pb emplace_back //emplace_backの方が速いが使い慣れてないため
#define pf push_front
#define MIN(a,b) a=min((a),(b))
#define MAX(a,b) a=max((a),(b))
#define SIZE(v) (int)v.size()
#define DOUBLE fixed << setprecision(15)
#define fi first
#define se second
#define mp make_pair
const double pi = acos(-1.0);
typedef vector<int> vi;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<char> vc;
typedef vector<double> vd;
typedef vector<bool> vb;
typedef deque<ll> dll;
typedef pair<ll, ll> P;
typedef vector<P> vP;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
ll dy[4] = { 1,0,-1,0 };
ll dx[4] = { 0,1,0,-1 };
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template< typename T >
struct edge {
	int src, to;
	T cost;

	edge(int to, T cost) : src(-1), to(to), cost(cost) {}

	edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

	edge &operator=(const int &x) {
		to = x;
		return *this;
	}

	operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WeightedGraph = vector< Edges< T > >;
using UnWeightedGraph = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;
bool out_check(ll a, ll b) {
	if (0 <= a && a < b) return true;
	else return false;
}
void addmod(ll &a, ll &b) {
	a = (a + b) % mod;
}
void Pvll(vll v) {
	puts("vll------------------------------------------------");
	rep(i, v.size()) cout << v[i] << " ";
	puts("");
	puts("------------------------------------------------");
}
void Pvvll(vvll v) {
	puts("vvll------------------------------------------------");
	rep(i, v.size()) {
		rep(j, v[i].size()) {
			cout << v[i][j] << " ";
		}
		cout << endl;
	}
	puts("------------------------------------------------");
}

void Pvs(vs s) {
	puts("vs------------------------------------------------");
	rep(i, s.size()) {
		rep(j, s[i].size()) {
			cout << s[i][j] << " ";
		}
		cout << endl;
	}
	puts("------------------------------------------------");
}

void Yes(bool x) {
	if (x) cout << "Yes\n";
	else cout << "No\n";
}

void YES(bool x) {
	if (x) cout << "YES\n";
	else cout << "NO\n";
}

void yes(bool x) {
	if (x) cout << "yes\n";
	else cout << "no\n";
}

void Yay(bool x) {
	if (x) cout << "Yay!\n";
	else cout << ":(\n";
}

ll  n, m, d, r, l, k, h, ans, ret = M;
bool flag = false, flag2 = false, flag3 = false;
//string s, t, u;

void INIT() {
	cin.tie(0);
	ios::sync_with_stdio(false);
}
int main() {
	INIT();//not intaractive
	ll g;
	cin >> d >> g;
	vll p(d), c(d);
	rep(i, d)cin >> p[i] >> c[i];
	ans = M;
	rep(bit, (1 << d)) {
		ll s = 0, num = 0, ma = -1;
		rep(i, d) {
			if (bit >> i & 1) {
				s += 100 * (i + 1)*p[i] + c[i];
				num += p[i];
			}
			else ma = i;
		}
		if (s < g) {
			ll s1 = 100 * (ma + 1);
			ll need = (g - s + s1 - 1) / s1;
			if (need >= p[ma]) continue;
			num += need;
		}
		chmin(ans, num);
	}
	cout << ans << endl;
	return 0;
}