codeforces A#264. Caisa and Sugar

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Caisa在超市购买巧克力蛋糕所需糖时，利用特殊换零政策，通过数学计算找到最优解决方案，最大化获取额外糖的数量。

Caisa is going to have a party and he needs to buy the ingredients for a big chocolate cake. For that he is going to the biggest supermarket in town.

Unfortunately, he has just s dollars for sugar. But that's not a reason to be sad, because there aren types of sugar in the supermarket, maybe he able to buy one. But that's not all. The supermarket has very unusual exchange politics: instead of cents the sellers give sweets to a buyer as a change. Of course, the number of given sweets always doesn't exceed 99, because each seller maximizes the number of dollars in the change (100 cents can be replaced with a dollar).

Caisa wants to buy only one type of sugar, also he wants to maximize the number of sweets in the change. What is the maximum number of sweets he can get? Note, that Caisa doesn't want to minimize the cost of the sugar, he only wants to get maximum number of sweets as change.

Input

The first line contains two space-separated integers n, s(1 ≤ n, s ≤ 100).

The i-th of the next n lines contains two integers x_i,y_i(1 ≤ x_i ≤ 100; 0 ≤ y_i < 100), wherex_i represents the number of dollars andy_i the number of cents needed in order to buy thei-th type of sugar.

Output

Print a single integer representing the maximum number of sweets he can buy, or-1 if he can't buy any type of sugar.

Sample test(s)

Input

Output

Input

Output

-1

Note

In the first test sample Caisa can buy the fourth type of sugar, in such a case he will take50 sweets as a change.

题意：给你n个糖果，s美元，每个糖果给出美元和美分，我们买一种的时候，找回来的零的美分是等价于同数值个数的糖果，问你能找回最多的糖果是多少

思路：判断就是了‘

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 205;

int n, s;
int x[maxn], y[maxn];

int main() {
	scanf("%d%d", &n, &s);
	int ans = -1;
	for (int i = 0; i < n; i++) {
		scanf("%d%d", &x[i], &y[i]);
		if (x[i] < s && y[i] != 0)
			ans = max(ans, 100 - y[i]);
		if (x[i] < s && y[i] == 0)
			ans = max(ans, 0);
		if (x[i] == s && y[i] == 0)
			ans = max(ans, 0);
	}
	printf("%d\n", ans);	
}