E. Change-free

Student Arseny likes to plan his life for n days ahead. He visits a canteen every day and he has already decided what he will order in each of the following n days. Prices in the canteen do not change and that means Arseny will spend c_i rubles during the i-th day.

There are 1-ruble coins and 100-ruble notes in circulation. At this moment, Arseny has m coins and a sufficiently large amount of notes (you can assume that he has an infinite amount of them). Arseny loves modern technologies, so he uses his credit card everywhere except the canteen, but he has to pay in cash in the canteen because it does not accept cards.

Cashier always asks the student to pay change-free. However, it's not always possible, but Arseny tries to minimize the dissatisfaction of the cashier. Cashier's dissatisfaction for each of the days is determined by the total amount of notes and coins in the change. To be precise, if the cashier gives Arseny x notes and coins on the i-th day, his dissatisfaction for this day equals x·w_i. Cashier always gives change using as little coins and notes as possible, he always has enough of them to be able to do this.

"Caution! Angry cashier"

Arseny wants to pay in such a way that the total dissatisfaction of the cashier for n days would be as small as possible. Help him to find out how he needs to pay in each of the n days!

Note that Arseny always has enough money to pay, because he has an infinite amount of notes. Arseny can use notes and coins he received in change during any of the following days.

Input

The first line contains two integers n and m (1 ≤ n ≤ 10⁵, 0 ≤ m ≤ 10⁹) — the amount of days Arseny planned his actions for and the amount of coins he currently has.

The second line contains a sequence of integers c₁, c₂, ..., c_n (1 ≤ c_i ≤ 10⁵) — the amounts of money in rubles which Arseny is going to spend for each of the following days.

The third line contains a sequence of integers w₁, w₂, ..., w_n (1 ≤ w_i ≤ 10⁵) — the cashier's dissatisfaction coefficients for each of the following days.

Output

In the first line print one integer — minimum possible total dissatisfaction of the cashier.

Then print n lines, the i-th of then should contain two numbers — the amount of notes and the amount of coins which Arseny should use to pay in the canteen on the i-th day.

Of course, the total amount of money Arseny gives to the casher in any of the days should be no less than the amount of money he has planned to spend. It also shouldn't exceed 10⁶ rubles: Arseny never carries large sums of money with him.

If there are multiple answers, print any of them.

Examples

Input

5 42
117 71 150 243 200
1 1 1 1 1

Output

79
1 17
1 0
2 0
2 43
2 0

Input

3 0
100 50 50
1 3 2

Output

150
1 0
1 0
0 50

Input

5 42
117 71 150 243 200
5 4 3 2 1

Output

230
1 17
1 0
1 50
3 0
2 0

发自内心的感觉要是我的水平是x，现在这个题的水平就是x+1，应该多做这样的题，我的水平才能变成x+1；

废话不多说，现在我们来讲一下这个问题：

假设我们现在的到x的最优解为sum，这个是最小的东西，然后我们添加一个thing，这个thing的带来的bad权值就是(100-a[i]）*w[i];

要是现在我还有money我他妈的肯定要买的对吗？ 那么现在我的value是不增加的，要是我没有钱没这个东西，假设我们前面的就是最优的解

那么我们怎么确保加不加这个东西啊，或者说加了这个东西要去掉或者增加前面的那些东西，这是我们要解决的两个问题，

问题1： 我们加不加这个东西，要是前面的最小bad权值<当前的bad权值我们一定要加上这个东西的，要是不加的话结果肯定不是最优的，那么我们现在   要做的就是决定去掉哪个东西，我们去掉的肯定是最小的bad权值对应的编号，这是很显然的，但是我们现在又要思考去掉了这个东西，我们剩下的钱还能不能买前面的其他的东西了，答案是肯定不能买的，这个地方逻辑水平还是很高的，

假设当前的thing需要的coin为x，前面的最小的bad去掉后，我们现在有的coin=100+原来的coin1;

coin1<x 要是等于的话就能买了。 买了x后我们还有100+coin1-x; 这个东西肯定是小于100的那么说之前不能买的东西贡献的值为y

就相当于100-它需要的coin  ,  y+coin2<x  现在我们有100+coin2+y-x;我们要是能买前面的东西的话，我们最少的钱要是100-y

100+coin2+y-x<100-y