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)

#define _CRT_SECURE_NO_WARNINGS
#include "bits/stdc++.h"
#include <random>
#include <unordered_map>
#include <unordered_set>
//#include <opencv2/core.hpp>
//#include <opencv2/highgui.hpp>
//#include <opencv2/imgproc.hpp>

using namespace std;

//呪文
#define DUMPOUT cerr
#define dump(...) DUMPOUT<<"  ";DUMPOUT<<#__VA_ARGS__<<" :["<<__LINE__<<":"<<__FUNCTION__<<"]"<<endl;DUMPOUT<<"    ";dump_func(__VA_ARGS__)

typedef unsigned uint; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef pair<double, double> pdd; typedef pair<string, string> pss;
template <typename _KTy, typename _Ty> ostream& operator << (ostream& ostr, const pair<_KTy, _Ty>& m) { ostr << "{" << m.first << ", " << m.second << "}"; return ostr; }
template <typename _KTy, typename _Ty> ostream& operator << (ostream& ostr, const map<_KTy, _Ty>& m) { if (m.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << *m.begin(); for (auto itr = ++m.begin(); itr != m.end(); itr++) { ostr << ", " << *itr; } ostr << "}"; return ostr; }
template <typename _KTy, typename _Ty> ostream& operator << (ostream& ostr, const unordered_map<_KTy, _Ty>& m) { if (m.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << *m.begin(); for (auto itr = ++m.begin(); itr != m.end(); itr++) { ostr << ", " << *itr; } ostr << "}"; return ostr; }
template <typename _Ty> ostream& operator << (ostream& ostr, const vector<_Ty>& v) { if (v.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << v.front(); for (auto itr = ++v.begin(); itr != v.end(); itr++) { ostr << ", " << *itr; }	ostr << "}"; return ostr; }
template <typename _Ty> ostream& operator << (ostream& ostr, const set<_Ty>& s) { if (s.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << *(s.begin()); for (auto itr = ++s.begin(); itr != s.end(); itr++) { ostr << ", " << *itr; }	ostr << "}"; return ostr; }
template <typename _Ty> ostream& operator << (ostream& ostr, const unordered_set<_Ty>& s) { if (s.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << *(s.begin()); for (auto itr = ++s.begin(); itr != s.end(); itr++) { ostr << ", " << *itr; }	ostr << "}"; return ostr; }
template <typename _Ty> ostream& operator << (ostream& ostr, const stack<_Ty>& s) { if (s.empty()) { ostr << "{ }"; return ostr; } stack<_Ty> t(s); ostr << "{" << t.top(); t.pop(); while (!t.empty()) { ostr << ", " << t.top(); t.pop(); } ostr << "}";	return ostr; }
template <typename _Ty> ostream& operator << (ostream& ostr, const list<_Ty>& l) { if (l.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << l.front(); for (auto itr = ++l.begin(); itr != l.end(); ++itr) { ostr << ", " << *itr; } ostr << "}"; return ostr; }
template <typename _KTy, typename _Ty> istream& operator >> (istream& istr, pair<_KTy, _Ty>& m) { istr >> m.first >> m.second; return istr; }
template <typename _Ty> istream& operator >> (istream& istr, vector<_Ty>& v) { for (size_t i = 0; i < v.size(); i++) istr >> v[i]; return istr; }
namespace aux { // print tuple
	template<typename Ty, unsigned N, unsigned L> struct tp { static void print(ostream& os, const Ty& v) { os << get<N>(v) << ", "; tp<Ty, N + 1, L>::print(os, v); } };
	template<typename Ty, unsigned N> struct tp<Ty, N, N> { static void print(ostream& os, const Ty& value) { os << get<N>(value); } };
}
template<typename... Tys> ostream& operator<<(ostream& os, const tuple<Tys...>& t) { os << "{"; aux::tp<tuple<Tys...>, 0, sizeof...(Tys)-1>::print(os, t); os << "}"; return os; }

template<typename A, size_t N, typename T> inline void Fill(A(&array)[N], const T &val) { std::fill((T*)array, (T*)(array + N), val); }

void dump_func() { DUMPOUT << endl; }
template <class Head, class... Tail> void dump_func(Head&& head, Tail&&... tail) { DUMPOUT << head; if (sizeof...(Tail) == 0) { DUMPOUT << " "; } else { DUMPOUT << ", "; } dump_func(std::move(tail)...); }

#define PI 3.14159265358979323846
#define EPS 1e-10
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define fake false



int main() {

	cin.tie(0);
	ios::sync_with_stdio(false);

	ll numProblems, targetScore;
	cin >> numProblems >> targetScore;
	vector<ll> p(numProblems), c(numProblems);
	for (int i = 0; i < numProblems; i++)
		cin >> p[i] >> c[i];

	ll numPatterns = (1LL << numProblems);

	ll ans = INT_MAX;
	for (ll i = 0; i < numPatterns; i++) {
		// コンプリートするかどうか
		vector<bool> complete;
		ll n = i;
		for (ll b = 0; b < numProblems; b++)
			complete.push_back((n >> b) & 1);

		ll solved = 0, currentScore = 0;
		for (ll b = 0; b < numProblems; b++) {
			if (complete[b]) {
				solved += p[b];
				currentScore += (b + 1) * 100 * p[b] + c[b];
			}
		}
		
		// 残りの問題を priority_queue に突っ込む
		priority_queue<ll> pq;
		for (ll b = 0; b < numProblems; b++) {
			if (complete[b])
				continue;
			for (ll j = 0; j < p[b] - 1; j++)
				pq.push((b + 1) * 100);
		}

		// targetScore 超えるまで解く
		while (!pq.empty() && currentScore < targetScore) {
			currentScore += pq.top(); pq.pop();
			solved++;
		}

		if (currentScore < targetScore)
			continue;

		ans = min(ans, solved);
	}

	cout << ans << endl;

	return 0;
}