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 <bits/stdc++.h>
#define int long long
#define lint long long
#define pii pair<int,int>
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(c) (c).begin(),(c).end()
#define ZERO(a) memset(a,0,sizeof(a))
#define MINUS(a) memset(a,0xff,sizeof(a))
#define MINF(a) memset(a,0x3f,sizeof(a))
#define POW(n) (1LL<<(n))
#define POPCNT(n) (__builtin_popcount(n))
#define IN(i,a,b) (a <= i && i <= b)
using namespace std;
template <typename T> inline bool CHMIN(T& a,T b) { if(a>b) { a=b; return 1; } return 0; }
template <typename T> inline bool CHMAX(T& a,T b) { if(a<b) { a=b; return 1; } return 0; }
template <typename T> inline void SORT(T& a) { sort(ALL(a)); }
template <typename T> inline void REV(T& a) { reverse(ALL(a)); }
template <typename T> inline void UNI(T& a) { sort(ALL(a)); a.erase(unique(ALL(a)),a.end()); }
template <typename T> inline T LB(vector<T>& v, T a) { return *lower_bound(ALL(v),a); }
template <typename T> inline int LBP(vector<T>& v, T a) { return lower_bound(ALL(v),a) - v.begin(); }
template <typename T> inline T UB(vector<T>& v, T a) { return *upper_bound(ALL(v),a); }
template <typename T> inline int UBP(vector<T>& v, T a) { return upper_bound(ALL(v),a) - v.begin(); }
template <typename T1, typename T2> ostream& operator<< (ostream& os, const pair<T1,T2>& p) { os << p.first << " " << p.second; return os; }
template <typename T1, typename T2> istream& operator>> (istream& is, pair<T1,T2>& p) { is >> p.first >> p.second; return is; }
template <typename T> ostream& operator<< (ostream& os, const vector<T>& v) { REP(i,v.size()) { if (i) os << " "; os << v[i]; } return os; }
template <typename T> istream& operator>> (istream& is, vector<T>& v) { for(T& in : v) is >> in; return is; }
template <typename T = int> vector<T> make_v(size_t a) { return vector<T>(a); }
template <typename T, typename... Ts> auto make_v(size_t a, Ts... ts) { return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...)); }
template <typename T, typename V> typename enable_if<is_class<T>::value == 0>::type fill_v(T &t, const V &v) { t = v; }
template <typename T, typename V> typename enable_if<is_class<T>::value != 0>::type fill_v(T &t, const V &v) { for(auto &e : t) fill_v(e,v); }
const lint MOD = 1000000007;
const lint INF = 0x3f3f3f3f3f3f3f3f;
const double EPS = 1e-10;

int D, G;
int p[11], c[11];

void _main() {
    cin >> D >> G;
    REP(i, D) cin >> p[i] >> c[i];
    int ans = INF;
    REP(ms, POW(D)) {
        int sum = 0;
        int cnt = 0;
        REP(i, D) {
            if (ms >> i & 1) {
                sum += c[i] + (i + 1) * 100 * p[i];
                cnt += p[i];
            }
        }
        int rem = G - sum;
        for (int i = D - 1; rem > 0 && i >= 0; i--) {
            if (!(ms >> i & 1)) {
                int k = (rem + (i + 1) * 100 - 1) / ((i + 1) * 100);
                if (k < p[i]) {
                    cnt += k;
                    rem -= k * (i + 1) * 100;
                    break;
                }
            }
        }
        if (rem <= 0) CHMIN(ans, cnt);
    }
    cout << ans << endl;
}

signed main(signed argc, char **argv) {
    if (argc > 1) {
        if (strchr(argv[1], 'i'))
            freopen("input.txt", "r", stdin);
        if (strchr(argv[1], 'o'))
            freopen("output.txt", "w", stdout);
    }
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(10);
    _main();
    return 0;
}