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Igor爱上Tanya，想要用油漆在她家对面的围栏上写下数字来表达爱意。面对有限的油漆和不同数字的消耗量，如何找出最大可能的数字成为了一道有趣的数学挑战。

Igor has fallen in love with Tanya. Now Igor wants to show his feelings and write a number on the fence opposite to Tanya’s house. Igor thinks that the larger the number is, the more chance to win Tanya’s heart he has.

Unfortunately, Igor could only get v liters of paint. He did the math and concluded that digit d requires ad liters of paint. Besides, Igor heard that Tanya doesn’t like zeroes. That’s why Igor won’t use them in his number.

Help Igor find the maximum number he can write on the fence.

Input
The first line contains a positive integer v (0 ≤ v ≤ 106). The second line contains nine positive integers a1, a2, …, a9 (1 ≤ ai ≤ 105).

Output
Print the maximum number Igor can write on the fence. If he has too little paint for any digit (so, he cannot write anything), print -1.

Examples
Input
5
5 4 3 2 1 2 3 4 5
Output
55555

Input
2
9 11 1 12 5 8 9 10 6
Output
33

Input
0
1 1 1 1 1 1 1 1 1
Output
-1

题意

有v升油漆，给出涂写数字1-9分别所需的油漆，求能写的最大数字

解题思路

读题好费劲，读了好几遍才懂，分析了一下样例发现都是使用油漆最少的数，很显然找使用油漆最小的数n能得到最大位数max，那得到的这个数就是最大的，直接输出max次n，交了一发之后wa了，又想了一下发现有特殊情况没考虑到，就是油漆不能整除所找的最小数时，也就是说油漆还有剩的，这时候的高位上的数有可能可以大于n；那就在n~9之间寻找符合的数。
代码实现

#include<iostream>
using namespace std;
int main()
{
	int v;
	int a[10];
	int min = 1000000;
	int t;
	cin >> v;
	for (int i = 1; i <= 9; i++)//最小数值
	{
		cin >> a[i];
		if (a[i] <= min)
		{
			min = a[i];
			t=i;
		}
	}
	if (v < min)//最小值大于v，无解
	{
		cout << "-1" << endl;
		return 0;
	}
	int max, k;
	max = v / min;  //最大位数
	k = v % min;//余数
	for (int i; k > 0 && max;) //使高位尽量大
	{   
		for (i = 9; i > t&& max; i--) 
		{
			if (k >= (a[i] - a[t])) 
			{
				cout<< i;
				max = max - 1;  
				k -= (a[i] - a[t]);
				break;
			}
		}
		if (i == t) break;
	}
	for (int i = 0; i < max; i++) 
		cout << t;
	return 0;
}