Codeforces Round #447 (Div. 2)

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　A. QAQ

"QAQ" is a word to denote an expression of crying. Imagine "Q" as eyes with tears and "A" as a mouth.

Now Diamond has given Bort a string consisting of only uppercase English letters of length n. There is a great number of "QAQ" in the string (Diamond is so cute!).

Bort wants to know how many subsequences "QAQ" are in the string Diamond has given. Note that the letters "QAQ" don‘t have to be consecutive, but the order of letters should be exact.

Input

The only line contains a string of length n (1?≤?n?≤?100). It‘s guaranteed that the string only contains uppercase English letters.

Output

Print a single integer — the number of subsequences "QAQ" in the string.

Examples

Input

QAQAQYSYIOIWIN

Output

4

Input

QAQQQZZYNOIWIN

Output

3

Note

In the first example there are 4 subsequences "QAQ": "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN", "QAQAQYSYIOIWIN".

题解：暴力枚举位置啦

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3
 4 char a[105];
 5
 6 int main()
 7 {   scanf("%s",a);
 8     int n=strlen(a),ans=0;
 9     for(int i=0;i<n;i++){
10         if(a[i]!=‘Q‘) continue;
11         for(int j=i+1;j<n;j++){
12             if(a[j]!=‘A‘) continue;
13             for(int k=j+1;k<n;k++) if(a[k]==‘Q‘) ans++;
14         }
15     }
16     cout<<ans<<endl;
17 } 

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　C. Marco and GCD Sequence

In a dream Marco met an elderly man with a pair of black glasses. The man told him the key to immortality and then disappeared with the wind of time.

When he woke up, he only remembered that the key was a sequence of positive integers of some length n, but forgot the exact sequence. Let the elements of the sequence be a₁,?a₂,?...,?a_n. He remembered that he calculated gcd(a_i,?a_i?+?1,?...,?a_j) for every 1?≤?i?≤?j?≤?n and put it into a set S. gcd here means the greatest common divisor.

Note that even if a number is put into the set S twice or more, it only appears once in the set.

Now Marco gives you the set S and asks you to help him figure out the initial sequence. If there are many solutions, print any of them. It is also possible that there are no sequences that produce the set S, in this case print -1.

Input

The first line contains a single integer m (1?≤?m?≤?1000) — the size of the set S.

The second line contains m integers s₁,?s₂,?...,?s_m (1?≤?s_i?≤?10⁶) — the elements of the set S. It‘s guaranteed that the elements of the set are given in strictly increasing order, that means s₁?<?s₂?<?...?<?s_m.

Output

If there is no solution, print a single line containing -1.

Otherwise, in the first line print a single integer n denoting the length of the sequence, n should not exceed 4000.

In the second line print n integers a₁,?a₂,?...,?a_n (1?≤?a_i?≤?10⁶) — the sequence.

We can show that if a solution exists, then there is a solution with n not exceeding 4000 and a_i not exceeding 10⁶.

If there are multiple solutions, print any of them.

Examples

Input

42 4 6 12

Output

34 6 12

Input

22 3

Output

-1

Note

In the first example 2?=?gcd(4,?6), the other elements from the set appear in the sequence, and we can show that there are no values different from 2, 4, 6 and 12 among gcd(a_i,?a_i?+?1,?...,?a_j) for every 1?≤?i?≤?j?≤?n.

题解：剽悍的构造题，可惜比赛的时候只想出了一半！

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3
 4 const int maxn=1005;
 5
 6 int n;
 7 int a[maxn];
 8
 9 int main()
10 {   cin>>n;
11     int i;
12     for(i=1;i<=n;i++) cin>>a[i];
13     for(i=1;i<=n;i++) if(a[i]%a[1]) break;
14     if(i<=n) puts("-1");
15     else{
16         printf("%d\n%d",2*n-1,a[1]);
17         for(int j=2;j<=n;j++) printf(" %d %d",a[1],a[j]);
18     }
19     return 0;
20 }