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
inputCopy
2 3
1 2
outputCopy
2
inputCopy
3 17
10 10 10
outputCopy
3
inputCopy
1 9
10
outputCopy
-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.
之前题目看懂，但题意没理解清楚，就是 they are saved for her 那句，原来是可以存储到后面。。。。
之前果断wa5。。。。
#include
using namespace std;
typedef long long ll;
#define inf 0x3f3f3f3f
#define maxn 2005
const int mod=1e9+7;
#define eps 1e-5
#define pi acos(-1.0)
ll quickpow(ll a,ll b)
{
ll ans=1;
while(b){
if(b&1){
ans=ans*a;
}
a=a*a;
b>>=1;
}
return ans;
}
ll gcd(ll a,ll b)
{
return b==0?a:gcd(b,a%b);
}
int a[105];
int main()
{
ios::sync_with_stdio(false);
int n,k;
cin>>n>>k;
int i,j;
int ans=0;
int flag=0;
int id;
int cnt=0;
for(i=1;i>a[i];
}
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