CodeForces 219C Color Stripe

Color Stripe

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 219C

Description

A colored stripe is represented by a horizontal row of n square cells, each cell is pained one of k colors. Your task is to repaint the minimum number of cells so that no two neighbouring cells are of the same color. You can use any color from 1 to k to repaint the cells.

Input

The first input line contains two integers n and k (1 ≤ n ≤ 5·10⁵; 2 ≤ k ≤ 26). The second line contains n uppercase English letters. Letter "A" stands for the first color, letter "B" stands for the second color and so on. The first k English letters may be used. Each letter represents the color of the corresponding cell of the stripe.

Output

Print a single integer — the required minimum number of repaintings. In the second line print any possible variant of the repainted stripe.

Sample Input

Input

6 3
ABBACC

Output

2
ABCACA

Input

3 2
BBB

Output

1
BAB

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 int main()
 {
   int n,m;
   char s[];
   while(scanf("%d%d",&n,&m)!=EOF)
   {
   scanf("%s",s);
   int i,ans=;
   m+=;
   if(n==)
  {
      cout<<ans<<endl<<s<<endl;
      continue;
  }
      if(m==)
      {
          int l=,r=;
          for(i=;i<n;i++)
          if(i%==)
          {
              if(s[i]=='B') l++;
              else r++;
          }
          else
          {
              if(s[i]=='A') l++;
              else r++;
          }
          if(l<r)
          {
              cout<<l<<endl;
              for(i=;i<n;i++)
              if(i%==) cout<<'A';
              else cout<<'B';
          }
          else
          {
              cout<<r<<endl;
              for(i=;i<n;i++)
              if(i%==) cout<<'B';
              else cout<<'A';
          }
          cout<<endl;
          continue;
      }
   s[n]='#';
    for(i=;i<n;i++)
   if(s[i]==s[i-])
   {
       ans++;
       for(int j=;j<=m;j++)
       if(j==s[i]||j==s[i+]) continue;
       else
       {
           s[i]=j;
           break;
       }
   }
   s[n]='\0';
   cout<<ans<<endl<<s<<endl;

  }
     return ;
 }