## Description

Ivan wants to make a necklace as a present to his beloved girl. A necklace is a cyclic sequence of beads of different colors. Ivan says that necklace is beautiful relative to the cut point between two adjacent beads, if the chain of beads remaining after this cut is a palindrome (reads the same forward and backward).

Ivan has beads of n colors. He wants to make a necklace, such that it's beautiful relative to as many cuts as possible. He certainly wants to use all the beads. Help him to make the most beautiful necklace.

• time limit per test: 2 seconds
• memory limit per test: 256 megabytes
• input: standard input
• output: standard output

## Input

The first line of the input contains a single number $n (1 ≤ n ≤ 26)$ — the number of colors of beads. The second line contains after n positive integers $a_i$ — the quantity of beads of $i$-th color. It is guaranteed that the sum of $a_i$ is at least 2 and does not exceed 100 000.

## Output

In the first line print a single number — the maximum number of beautiful cuts that a necklace composed from given beads may have. In the second line print any example of such necklace.

Each color of the beads should be represented by the corresponding lowercase English letter (starting with a). As the necklace is cyclic, print it starting from any point.

## Examples

Input #1

3
4 2 1


Output #1

1
abacaba


Input #2

1
4


Output #2

4
aaaa


Input #3

2
1 1


Output #3

0
ab


## Note

In the first sample a necklace can have at most one beautiful cut. The example of such a necklace is shown on the picture.

In the second sample there is only one way to compose a necklace.

## Analysis

• 有一个奇数：每段都是回文的，这个奇数的字母必然放在每段中间；
• 全偶数：每段任意，只要关于切割点对称即可。

## Code

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>

using namespace std;
#define int long long

int ret=0,f=1;char ch=getchar();
while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
while (ch>='0'&&ch<='9') ret=ret*10+ch-'0',ch=getchar();
return ret*f;
}

const int maxn=30;

int n;
int a[maxn];

inline void imp(){
printf("0\n");
for (int i=1;i<=n;i++){
for (int j=1;j<=a[i];j++) putchar(i+'a'-1);
}
printf("\n");
}

inline int gcd(int x,int y){
return y?gcd(y,x%y):x;
}

signed main(){
for (int i=1;i<=n;i++){
if ((a[i]&1) && flg) no=true;
if (a[i]&1) flg=true;
}
if (no){
imp();
return 0;
}
if (n==1){
printf("%lld\n",a[1]);
for (int i=1;i<=a[1];i++) putchar('a');
printf("\n");
return 0;
}

int g;
g=a[1];
for (int i=2;i<=n;i++) g=gcd(g,a[i]);
if (!flg){
printf("%lld\n",g);
string otp1="",opt2="";
for (int i=1;i<=n;i++){
for (int j=1;j<=a[i]/g;j++) otp1+=(char)(i+'a'-1);
}
for (int i=n;i>=1;i--){
for (int j=1;j<=a[i]/g;j++) opt2+=(char)(i+'a'-1);
}
for (int i=1;i<=g;i++) if (i&1) cout<<otp1; else cout<<opt2;
printf("\n");
} else {
printf("%lld\n",g);
string opt="";int mid;
for (int i=1;i<=n;i++) if (a[i]&1) {mid=i;break;}
for (int i=1;i<=n;i++) if (i!=mid){
for (int j=1;j<=a[i]/g/2;j++) opt+=(char)(i+'a'-1);
}
for (int i=1;i<=a[mid]/g;i++) opt+=(char)(mid+'a'-1);
for (int i=n;i>=1;i--) if (i!=mid){
for (int j=1;j<=a[i]/g/2;j++) opt+=(char)(i+'a'-1);
}
for (int i=1;i<=g;i++) cout<<opt;
printf("\n");
}
return 0;
}