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)

/**
 *    Author:  "Robin Singh" !!!!    
**/
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(),(x).end()
#define ff first
#define ss second
#define mp make_pair
#define str string 
#define szz(v) ((int)(v).size())  
#define fr(i, j, k) for(int i = j; i < k; i++)     // 0 -> n-1 (i,0,n)
#define bk(i, j, k) for(int i = j-1; i >= k; i--)  // n-1 -> 0 (i,n,0)
#define frit(it, a) for(__typeof((a).begin()) it = a.begin(); it != a.end(); ++it)
#define mem(ptr)  memset(ptr, 0, sizeof ptr)  // use fill() for value other than 0 //
#define rtn return 
#define itt(a)   __typeof((a).begin())
#define RT  (cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << " ms\n")  // Run Time
#define nl "\n"  // New Line
#define dbg(x) (cerr << #x << " = " << x << "  ")
#define enl (cerr << "\n")
#define dbgg(x,y) (cerr << #x << " = " << x << " -> " << #y << " = " << y << "\n")
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<long long> vll;
typedef vector< pair<int,int> > vpii;
typedef vector< pair<ll, ll> > vpll;
const long long mod = 1e9+7;
const ld eps = 1e-9;
const int inf = (1<<30);
const ll INF = (1LL<<61);

ll gcd(ll a,ll b) { return b==0 ? a : gcd(b, a%b); }
ll expo(ll x, ll y){ ll res = 1; x = x % mod;  while (y > 0) { if (y & 1)res = (res*x) % mod; y = y>>1;x = (x*x) % mod;}return res;}
void matmul(ll a[][2], ll b[][2]) { ll tmp[2][2]; for(int i=0; i<2; i++) { for(int j=0; j<2; j++) { tmp[i][j] = 0; for(int k=0; k<2; k++) tmp[i][j] = (tmp[i][j] + a[i][k]*b[k][j]) % mod; }} for(int i=0; i<2; i++) for(int j=0; j<2; j++) a[i][j] = tmp[i][j]; }  // answer stored in matrix A
ll fibo(int n)    // returns nth fibonacci number, also S(n) = f(n+2)-1
{ if(n == 0) return 0; else if(n == 1) return 1; ll T[2][2] = { {0, 1} , {1, 1} }; ll res[2][2] = { {1, 0}, {0,1} }; n--; while(n) { if(n & 1) matmul(res, T); n >>= 1; matmul(T, T); } return res[1][1]; }
ll ainv, binv;
void extgcd(ll a, ll b){ if(b==0){/* hcf = a; */ ainv = 1; binv = 0; } else{ extgcd(b, a%b); ll temp = ainv; ainv = binv; binv = temp - (a/b)*binv; } }

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    int d, g;
    cin >> d >> g;
    int p[d+1], b[d+1];
    fr(i, 1, d+1){
        cin >> p[i] >> b[i];
    }
    ll ans = 0, tot = g;
    ll res = INF;
    for(int i=d; i>=1; i--){
        ll t = 100*i*p[i] + b[i];
        if(g >= t){
            g -= t;
            ans += p[i];
        }
        else if(g < t){
            ll x = g/(100*i);
            if(x * 100 * i < g) x++;
            if(x >= p[i]) x = p[i];
            ans += x;
            res = min(ans, res);
            g = tot;
            ans = 0;
        }
    }
    
    vector < pair <ll, ll> > v;
    for(int i=1; i<=d; i++){
        v.pb({p[i], i});
    }
    sort(all(v));
    
    for(int i=0; i<d; i++){
       // dbgg(v[i].ff, v[i].ss);
        ll t = 100 * v[i].ss * v[i].ff + b[v[i].ss];
        if(g >= t){
            g -= t;
            ans += p[i];
        }
        else if(g < t){
            ll x = g/(100*v[i].ss);
            if(x * 100 * v[i].ss < g) x++;
            if(x >= v[i].ff) x = v[i].ff;
            ans += x;
            res = min(ans, res);
            g = tot;
            ans = 0;
        }
    }
    cout << res;
    return 0;
}