Codeforces Round #407 (Div. 1) C. The Great Mixing(bfs)

Sasha and Kolya decided to get drunk with Coke, again. This time they have k types of Coke. i-th type is characterised by its carbon dioxide concentration . Today, on the party in honour of Sergiy of Vancouver they decided to prepare a glass of Coke with carbon dioxide concentration . The drink should also be tasty, so the glass can contain only integer number of liters of each Coke type (some types can be not presented in the glass). Also, they want to minimize the total volume of Coke in the glass.

Carbon dioxide concentration is defined as the volume of carbone dioxide in the Coke divided by the total volume of Coke. When you mix two Cokes, the volume of carbon dioxide sums up, and the total volume of Coke sums up as well.

Help them, find the minimal natural number of liters needed to create a glass with carbon dioxide concentration . Assume that the friends have unlimited amount of each Coke type.

Input

The first line contains two integers n, k (0 ≤ n ≤ 1000, 1 ≤ k ≤ 106) — carbon dioxide concentration the friends want and the number of Coke types.

The second line contains k integers a1, a2, ..., ak (0 ≤ ai ≤ 1000) — carbon dioxide concentration of each type of Coke. Some Coke types can have same concentration.

Output

Print the minimal natural number of liter needed to prepare a glass with carbon dioxide concentration , or -1 if it is impossible.

Examples

input

400 4100 300 450 500

output

2

input

50 2100 25

output

3

Note

In the first sample case, we can achieve concentration  using one liter of Coke of types  and : .

In the second case, we can achieve concentration  using two liters of  type and one liter of  type: 

题意：给你k种浓度的饮料(ai/L),让你兑出来n/L浓度的饮料,每种饮料只能取整数升,为至少要多少升饮料.

分析:把所有的ai都减去n,问题就转化为了选择最少的数来凑零,bfs就行了,注意中间状态只需要开到ai的最大值就可以了.

#include <bits/stdc++.h>#define N 1005#define INF 2147483647#define mask 1002using namespace std;int k,n,a[N*N],q[N*N];int vis[2*N],f[2*N];int main(){cin.sync_with_stdio(false);cin>>n>>k;for(int i = 1;i <= k;i++) cin>>a[i];sort(a+1,a+1+k);k = unique(a+1,a+1+k) - a - 1;for(int i = 1;i <= k;i++) a[i] -= n;if(a[1]*a[k] > 0){cout<<-1<<endl;return 0;}int s = 1,t = 1;for(int i = 1;i <= k;i++) {q[t++] = a[i];f[a[i]+mask] = 1;vis[a[i]+mask] = true;}while(s != t && !vis[mask]){for(int i = 1;i <= k;i++) if(abs(a[i] + q[s]) <= 1002 && !vis[a[i] + q[s] + mask]) { f[a[i] + q[s] + mask] = f[q[s] + mask] + 1; vis[a[i] + q[s] + mask] = true; q[t++] = a[i] + q[s];  } s++; }cout<<f[mask]<<endl;}