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 <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <map>
#include <set>
#include <vector>
#include <functional>
#include <algorithm>
#include <stack>
#include <queue>
//#include <iostream>
//#include <sstream>
using namespace std;
#define all(cont) cont.begin(), cont.end()
#define rall(cont) cont.end(), cont.begin()
#define mp make_pair
#define pb push_back
#define fs first
#define sc second
#define pf printf
#define sf scanf
#define fr(i, a, b) for(int i = a ; i < b ; i++)
#define frr(i, a, b) for(int i = a ; i >= b ; i--)
#define tr(it, var) for(__typeof(var.begin()) it = var.begin(); it != var.end(); it++)
#define trr(rit, var) for(__typeof(var.rbegin()) rit = var.rbegin(); rit != var.rend(); rit++)
#define its(x) static_cast< ostringstream & >( ( ostringstream() << dec << x ) ).str()
#define boost ios_base::sync_with_stdio(false); cin.tie(NULL)
#define deb(x) cout << #x" = " << x << '\n'
#define has(cont, x) ((cont).find((x)) != (cont).end())
#define sz(x) ((int)(x).size())
//typedef
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
//bitmask
int onbit(int x, int pos) {x |= (1 << pos); return pos;}
int offbit(int x, int pos) {x &= ~(1 << pos); return x;}
int test(int x, int pos) {return x & (1 << pos);}
int flip(int x, int pos) {x ^= (1 << pos); return x;}
int lsb(int x) {return x & (-x);}
int onbitall(int x, int pos) {x = (1 << pos)-1; return x;}
//constants
const double EPS = 1e-9;
const double PI = 2*acos(0.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9+7;
//moves
const int dx1[] = {0, 1, 0, -1};
const int dy1[] = {1, 0, -1, 0};
const int dx2[] = {-1, -1, -1, 0, 0, 1, 1, 1};
const int dy2[] = {-1, 0, 1, -1, 1, -1, 0, 1};

int dp[1234567];

int main() {
//    freopen("input.in", "r", stdin);
	int d, g;
	sf("%d %d", &d, &g);
	pii a[1234];
	fr(i, 0, d) 
		sf("%d %d", &a[i].fs, &a[i].sc);
	fr(i, 0, 1234567) dp[i] = INF;
	frr(i, d-1, 0) {
		int score = i+1, cur = 1, start = 1;
		frr(h, i, 0) {
			for(int j = start ; j < start+a[h].fs*(h+1) ; j += (h+1)) {
				fr(k, cur, j+(h+1)) dp[k] = min(dp[k], cur);
				cur++;
			}
			for(int j = a[i].fs*(h+1) + a[i].sc/100 ; j < start+a[h].fs*(h+1)+a[i].fs*(h+1) + a[i].sc/100 ; j += (h+1)) {
				fr(k, cur, j+(h+1)) dp[k] = min(dp[k], cur);
				cur++;
			}
			start += a[h].fs*(h+1);
		}
//		pf("%d\n", a[i].fs*score + a[i].sc/100);
		dp[a[i].fs*score + a[i].sc/100] = min(dp[a[i].fs*score + a[i].sc/100], a[i].fs);
//		fr(i, 1, 30) pf("%d ", dp[i]); puts("");
	}
	int res = INF;
	fr(i, g/100, 12345) res = min(res, dp[i]);
	pf("%d\n", res);
	return 0;
}