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 <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <assert.h>
#include <numeric>
#include <math.h>
#include <climits>
#include <stack>
#include <iomanip>
#include <sstream>

using namespace std;

template <typename T>
inline bool sign(T A) {
    return (A>0)-(A<0);
}
#define REP(i, n) for (int i = 0; i < (n); ++i)
#define SREP(i,s,n) for(int i=s; (n-i)*sign(n-s)>0; i+=sign(n-s))
#define all(x) (x).begin(),(x).end()
typedef long long ll;
typedef unsigned long long ull;

template <typename T>
T gcd(T a, T b) {
    if(a < b) gcd(b, a);
    if (b == 1) return 1;
    T r;
    while ((r=a%b)) {
        a = b;
        b = r;
    }
    return b;
}

bool _less(pair<int, int> a, pair<int, int> b) {
    return a.second < b.second;
}

template <template<class,class,class...> class C, typename K, typename V, typename... Args>
V map_get(const C<K,V,Args...>& m, K const& key, const V & defval)
{
    typename C<K,V,Args...>::const_iterator it = m.find( key );
    if (it == m.end())
        return defval;
    return it->second;
}

#define MOD (1'000'000'000 + 7
#define INF (ll(1e18))

int main(void)
{
    ll D, G;
    cin >> D >> G;
    vector<ll> p(D), c(D);
    REP(i, D) {
        cin >> p[i] >> c[i];
    }
    const ll N = 200000;
    vector<vector<ll>> dp(D+1, vector<ll>(N, INF));
    dp[0][0] = 0;
    REP(i, D) {
        dp[i+1][0] = 0;
        REP(k, G/100)
        {
            if (dp[i][k] < INF)
            {
                REP(j, p[i]+1)
                {
                    ll s = (i + 1) * j;
                    if (j == p[i])
                    {
                        s += c[i] / 100;
                    }
                    dp[i + 1][k + s] = min(dp[i][k + s], min(dp[i + 1][k + s], dp[i][k] + j));
                }
            }
        }
    }
    ll score = INF;
    SREP(i, G/100, N) {
        score = min(dp[D][i], score);
    }
    cout << score << endl;
}