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 <bits/stdc++.h>
#include <regex>
#ifdef _DEBUG
#include "debug.h"
#endif
#define int ll
#define FOR(i, s, e) for (int i = int(s); i < int(e); ++i)
#define REP(i, e) FOR(i, 0, e)
#define all(v) (v).begin(),(v).end()
#define mkpr make_pair
#define mktp make_tuple
#define tv(tp, i) get<(i)>(tp)
#define CLEAR(obj) obj = decltype(obj)()
#define var auto
#define OUT(type) type &
#define IN(type) const type &
#define self (*this)
#define in :
#define True true
#define False false
using namespace std; using int32 = int32_t; using str = string; using cstr = const str;
using ll = long long; using ull = unsigned long long; using cint = const int; using cll = const ll;
using vi = vector<int>; using vvi = vector<vi>; using vl = vector<ll>; using vvl = vector<vl>;
using vb = vector<bool>; using vvb = vector<vb>; using vd = vector<double>; using vvd = vector<vd>;
using vc = vector<char>; using vvc = vector<vc>; using vs = vector<string>;
using pii = pair<int, int>; using pll = pair<ll, ll>; using pdd = pair<double, double>; using psl = pair<string, ll>;
using tiii = tuple<int, int, int>; using tlll = tuple<ll, ll, ll>;
static const int MGN = 10; static const int ARYSZ_MAX = (int)1e7;
static const ll LINF = LLONG_MAX/2;
#ifdef int
static const int INF = LINF;
#else
static const int INF = INT_MAX/2;
#endif
static double EPS = 1e-9;
#define FASTCIN() cin.tie(0);ios::sync_with_stdio(false);
#ifdef IMPORT_SAMPLE
#define DEBUG_IMPORT(file) vs args(argv, argv+argc);str sample = make_sample(args, file);ifstream ifs(sample); cin.rdbuf(ifs.rdbuf());
#else
#define DEBUG_IMPORT(file)
#endif
static inline bool DBLEQ(double a, double b) {return abs(a - b) < EPS;}
static inline void EPR(str msg) {cerr << msg << endl;}
static inline void AST(bool exp, str msg) {if(!exp){EPR(msg);} assert(exp);}
static inline void TAST(bool exp, str msg) {if(!exp){EPR(msg);} while(!exp){}}

int D,G;
vi P,C;
int ans;

void dfs(int d, int cnt, int res) {
    if (cnt > ans)
        return;

    if (d < 0) {
        if (res >= G) {
            ans = min(ans, cnt);
        }
        return;
    }

    REP(i, P[d]+1) {
        if (i == P[d])
            dfs(d-1, cnt+i, res + 100*(d+1)*i + C[d]);
        else
            dfs(d-1, cnt+i, res + 100*(d+1)*i);
    }
}


int32 main(int32 argc, char* argv[]) {
    FASTCIN();
    DEBUG_IMPORT("sample-1");


    cin>>D>>G;
    P = vi(D); C = vi(D);
    REP(i,D)
        cin>>P[i]>>C[i];

    ans = INF;
    dfs(D-1, 0, 0);

    cout << ans << endl;


    return 0;
}