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
 
// loop
#define FOR(i, a, b) for (int i = (a); i < (b); i++) 	// a ~ b-1	(ascending)
#define REP(i, n) FOR(i, 0, n)							// 0 ~ n-1
#define NREP(i, n) FOR(i, 1, n + 1)						// 1 ~ n
#define RFOR(i, a, b) for (int i = (a); i >= (b); i--)	// a ~ b	(descending)
#define RREP(i, n) RFOR(i, n, 0)						// n ~ 0
#define RNREP(i, n) RFOR(i, n, 1)						// n ~ 1
 
// container operation
#define EACH(i, c) for (auto i = (c).begin(); i != (c).end(); i++)
#define ASORT(c) std::sort((c).begin(), (c).end())
#define DSORT(c) std::sort((c).begin(), (c).end(), std::greater<typeof((c).front())>())
#define SIZE(x) ((int)(x).size())

// chmax, chmin
template <typename T>
bool chmax(T &a, const T& b) {
  if (a < b) {
    a = b;  // aをbで更新
    return true;
  }
  return false;
}
template <typename T>
bool chmin(T &a, const T& b) {
  if (a > b) {
    a = b;  // aをbで更新
    return true;
  }
  return false;
}

// debug
#define check(x) std::cout << #x << " = " << x << '\n'
 
// print
#define cout(x) std::cout << x << '\n'
 
// type alias
using VI = std::vector<int>;
using VII = std::vector<VI>;
using VB = std::vector<bool>;
using VBB = std::vector<VB>;
using VS = std::vector<std::string>;
using PII = std::pair<int, int>;

std::string yes = "Yes\n";
std::string no = "No\n";

void solve();

signed main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  solve();
  return 0;
}

// [解法1] bit全探索
// void solve() {
//   int d, g;
//   std::cin >> d >> g;
//   std::vector<int> p(d), c(d);
//   REP(i, d) {
//     std::cin >> p.at(i) >> c.at(i);
//   }
//
//   int min_num = 100000;
//   REP(i, (1 << d)) {
//     std::bitset<10> b(i);
//     int score = 0;
//     int num = 0;
//
//     // Total score of completed problem sets and the number of problems
//     REP(j, d) {
//       if (b.test(j)) {
//         score += 100 * (j+1) * p.at(j) + c.at(j);
//         num += p.at(j);
//       }
//     }
//
//     // Calculate the problem needs to be solved
//     if (g > score) {
//       // Find the problem set with the highest points from incomplete problem sets
//       int h_score;
//       int h_num;
//       for (int j = d-1; j >= 0; j--) {
//         if (!b.test(j)) {
//           h_score = 100 * (j + 1);
//           h_num = p.at(j);
//           break;
//         }
//       }
//       if (g - score > (h_num - 1) * h_score) {
//         continue;
//       }
//       num += (g - score + h_score - 1) / h_score;
//     }
//
//     // update min
//     if (num < min_num) {
//       min_num = num;
//     }
//   }
//   cout(min_num);
// }

// [解法2] 左から決めていく再帰による深さ優先探索
int d, g;
std::vector<int> p(10), c(10);
int min = 100000;
void dfs(int i, std::set<int> s, int totals, int totaln);
void solve() {
  std::cin >> d >> g;
  std::set<int> s;
  REP(i, d) {
    std::cin >> p.at(i) >> c.at(i);
    s.insert(i);
  }
  dfs(0, s, 0, 0);
  cout(min);
}
void dfs(int i, std::set<int> s, int totals, int totaln) {
  if (i == d) {
    if (s.empty()) {
      if (totals >= g) chmin(min, totaln);
      return;
    }
    // 全て解かない物の中で最も点数の高いものを選ぶ
    int maxi = *s.rbegin();
    if (totals + 100 * (maxi+1) * (p.at(maxi) - 1) < g) return;
    chmin(min, totaln + std::max((int)0, ((g - totals) + 100 * (maxi+1) - 1) / (100 * (maxi+1))));
    return;
  }
  // i番目を全て解かない
  dfs(i+1, s, totals, totaln);
  // i番目を全て解く
  s.erase(i);
  dfs(i+1, s, totals + 100*(i+1)*p.at(i) + c.at(i), totaln + p.at(i));
}