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).

Examples

Input

2 700
3 500
5 800


Output

3


Input

2 2000
3 500
5 800


Output

7


Input

2 400
3 500
5 800


Output

2


Input

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000


Output

66
#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for (int i = 0; i < (n); ++i)
typedef long long ll;
const int INF=1e9+7;

int main() {
	ll D,G,r=INF;
	cin >> D >> G;
	vector<ll> p(D),c(D),s(D);

	rep(i,D) {
		cin >> p.at(i) >> c.at(i);
		s.at(i) = p.at(i)*(i+1)*100 + c.at(i);
	}
	
	rep(i,(int)pow(2,D)-1) {
		ll tq=0,ts=0,m=-1;
		rep(j,D)
			if(i>>j & 1) {
				tq += p.at(j);
				ts += s.at(j);
			} else
				m = j;

		if(ts >= G) {
			r = min(r,tq);
			continue;
		}

		rep(i,p.at(m)) {
			++tq;
			ts += (m+1)*100;
			if(i == p.at(m)-1)
				ts += c.at(m);
			if(ts >= G) {
				r = min(r,tq);
				break;
			}
		}
	}

	cout << r << endl;
}