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 <bits/stdc++.h>
using namespace std;
#define loop(i,b,e) for(int i=b;i<=e;i++)
#define loop2(i,e,b) for(int i=e;i>=b;i--)
#define io ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define fi first
#define se second
#define X real()
#define Y imag()
#define cp(a,b)	( (conj(a)*(b)).imag() )
#define length(a) (hypot((a).imag(), (a).real()))
typedef complex<double> point;
typedef long long ll;
typedef unsigned long long ull;
double PI = 3.141592653589793;
const double EPS = 1e-9;
const int N = 1e6 + 5;
const ll mod = 1e9 + 7;
const int oo = 1e9;
int dx[] = { 0, 0, 1, -1, 1, -1, 1, -1 };
int dy[] = { 1, -1, 0, 0, -1, 1, 1, -1 };

int d;
pair<ll, ll> t[N];
ll g;
ll ans = 1e18;
bool vis[105];

void solve(ll sum,ll solved) {
	if (sum >= g) {
		ans = min(solved, ans);
		return;
	}
	for (int i = 0; i < d; i++) {
		if (vis[i]) continue;
		if (sum + t[i].first * (i + 1) * 100 > g) {
			ll need = g - sum;
			ans = min(ans,solved + ( need / ((i+ 1) * 100)) );
			continue;
		}
		vis[i] = 1;
		solve(sum + t[i].first * (i + 1) * 100 + t[i].second, solved + t[i].first);
		vis[i] = 0;
	}
}

int main() {
	scanf("%d%lld",&d,&g);
	loop(i, 0, d - 1) {
		scanf("%lld%lld",&t[i].first,&t[i].second);
	}
	solve(0, 0);
	cout << ans;
}