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)

/*بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيم*/
#include <bits/stdc++.h>
using namespace std;
#define INF 1<<30
#define endl '\n'
#define maxn 1000005
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
typedef long long ll;
const double PI = acos(-1.0);
const ll mod = 1e9 + 7;

inline void normal(ll &a) { a %= mod; (a < 0) && (a += mod); }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
inline ll modSub(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); a -= b; normal(a); return a; }
inline ll modPow(ll b, ll p) { ll r = 1; while (p) { if (p & 1) r = modMul(r, b); b = modMul(b, b); p >>= 1; } return r; }
inline ll modInverse(ll a) { return modPow(a, mod - 2); }
inline ll modDiv(ll a, ll b) { return modMul(a, modInverse(b)); }


/* for Random Number generate
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
*/
///**
template < typename F, typename S >
ostream& operator << ( ostream& os, const pair< F, S > & p ) {
  return os << "(" << p.first << ", " << p.second << ")";
}

template < typename T >
ostream &operator << ( ostream & os, const vector< T > &v ) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if ( it != v.begin() ) os << ", ";
    os << *it;
  }
  return os << "}";
}

template < typename T >
ostream &operator << ( ostream & os, const set< T > &v ) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if ( it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}

template < typename F, typename S >
ostream &operator << ( ostream & os, const map< F, S > &v ) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if ( it != v.begin() ) os << ", ";
    os << it -> first << " = " << it -> second ;
  }
  return os << "]";
}

#define dbg(args...) do {cerr << #args << " : "; faltu(args); } while(0)

clock_t tStart = clock();
#define timeStamp dbg("Execution Time: ", (double)(clock() - tStart)/CLOCKS_PER_SEC)

void faltu () { cerr << endl; }

template <typename T>
void faltu( T a[], int n ) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << endl;
}

template <typename T, typename ... hello>
void faltu( T arg, const hello &... rest) { cerr << arg << ' '; faltu(rest...); }
// Program showing a policy-based data structure.
#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp>
#include <functional> // for less 
#include <iostream>
using namespace __gnu_pbds;
using namespace std;
// GNU link : https://goo.gl/WVDL6g
typedef tree<int, null_type, less_equal<int>, rb_tree_tag,
        tree_order_statistics_node_update>
        new_data_set;

// find_by_order(k) – ফাংশনটি kth ordered element এর একটা পয়েন্টার রিটার্ন করে। অর্থাৎ তুমি চাইলেই kth ইন্ডেক্সে কি আছে, সেটা জেনে ফেলতে পারছো!
// order_of_key(x) – ফাংশনটি x এলিমেন্টটা কোন পজিশনে আছে সেটা বলে দেয়।

//*//**___________________________________________________**/



int main()
{
  FASTIO
  /*
#ifndef ONLINE_JUDGE
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
  freopen("error.txt", "w", stderr);
#endif
//*/
  ll n, G;
  cin >> n >> G;
  vector<pair<int, int> > vec;
  for (int i = 0; i < n; i++) {
    ll p, c;
    cin >> p >> c;
    vec.push_back({p, c});
  }
  //dbg(vec);
  ll ans = INT_MAX;
  for (int i = 0; i < (1 << n); i++) {
    ll res = 0ll;
    ll cnt = 0;
    for (int j = 0; j < n; j++) {
      if ((1 << j)&i) {
        ll score = vec[j].first * (j + 1) * 100;
        ll bonus = vec[j].second;
        cnt += vec[j].first;
        res += score + bonus;
      }
    }
    //dbg(res, cnt);
    for (int t = n - 1; t >= 0; t--) {
      if ((1 << t)&i)continue;
      if (res >= G)break;
      ll ase = vec[t].first;
      ll score = (t + 1) * 100;
      ll need = G - res;
      ll lagbe = (need) / score;
      ll mn = min(ase, lagbe);
      cnt += mn;
      res += mn * 100 * (t + 1);
    }
    if (res >= G)
      ans = min(ans, cnt);
  }
  cout << ans << endl;
  return 0;
}