codeforces 839A 之 Arya and Bran

A. Arya and Bran

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Bran and his older sister Arya are from the same house. Bran like candies so much, so Arya is going to give him some Candies.

At first, Arya and Bran have 0 Candies. There are n days, at the i-th day, Arya finds ai candies in a box, that is given by the Many-Faced God. Every day she can give Bran at most 8 of her candies. If she don't give him the candies at the same day, they are saved for her and she can give them to him later.

Your task is to find the minimum number of days Arya needs to give Bran k candies before the end of the n-th day. Formally, you need to output the minimum day index to the end of which k candies will be given out (the days are indexed from 1 to n).

Print -1 if she can't give him k candies during n given days.

Input

The first line contains two integers n and k (1 ≤ n ≤ 100, 1 ≤ k ≤ 10000).

The second line contains n integers a1, a2, a3, ..., an (1 ≤ ai ≤ 100).

Output

If it is impossible for Arya to give Bran k candies within n days, print -1.

Otherwise print a single integer — the minimum number of days Arya needs to give Bran k candies before the end of the n-th day.

Examples

input

2 31 2

output

2

input

3 1710 10 10

output

3

input

1 910

output

-1

Note

In the first sample, Arya can give Bran 3 candies in 2 days.

In the second sample, Arya can give Bran 17 candies in 3 days, because she can give him at most 8 candies per day.

In the third sample, Arya can't give Bran 9 candies, because she can give him at most 8 candies per day and she must give him the candies within 1 day.

题意：Arya 每天可以获得ai颗糖，她要给Bran    k颗糖，他每天最多给8颗，当天多余的糖可以留着以后给，问最少要第几天才可以满足要求？

AC代码如下：

#include <iostream>#include <cstdio>using namespace std;const int maxn=10000+10;int a[maxn];int main(){    int n,k;    while(cin>>n>>k)    {        int ans=0;        for(int i=0;i<n;i++)            cin>>a[i];        for(int i=0;i<n;i++)        {            if(a[i]<8)            {                if(k>=a[i])                {                    k-=a[i];                    if(k==0)                    {                         ans=i+1;                        break;                    }                }                else                {                    k=0;                    ans=i+1;                    break;                }            }            else            {                if(k<=8)                {                    k=0;                    ans=i+1;                    break;                }                else                {                    k-=8;                    a[i]-=8;                    a[i+1]+=a[i];                    if(k==0)                    {                         ans=i+1;                        break;                    }                }            }        }        if(!ans) cout<<-1<<endl;        else cout<<ans<<endl;    }    return 0;}