#include <cstdio>
#include <queue>

using namespace std;

int fact[]={1,1,2,6,24,120,720,5040};

int zakoduj(int perm[8])
{
   int pouz[9] = {0};
      
   int kod = 0, i, j;
   for (i = 0; i < 8; i++) {
      for (j = 1; j < perm[i]; j++)
         if (!pouz[j]) {
            kod += fact[7-i];
         }
      pouz[perm[i]] = 1;   
   }
   return kod;
}

void dekoduj(int kod, int perm[8])
{
   int pouz[9] = {0}, i, j, k;
   
   for (i = 0; i < 8; i++) {
      j = kod / fact[7-i];
      kod %= fact[7-i];
      for (k = 1; k <= 8; k++) {
         if (!pouz[k]) {
            if (!j)
               break;
            else
               j--;
         }
      }
      pouz[k] = 1;
      perm[i] = k;
   }
}


int main(void) {

   int perm[8], cil, permn[3][8], cur, kod, i, j;
   int hotovo[50000] = {0}, prev[50000] = {0};
   char path[50000], what[50000];
   FILE *vstup = fopen("obdelnik.in", "r"), *vystup;

   for (i = 0; i < 8; i++)
      fscanf(vstup, "%d", &(perm[i]));
   fclose(vstup);

   cil=zakoduj(perm);
   queue<int> q; // fronta

   hotovo[0] = 1;
   q.push(0); // 0 je v našem zápisu startovní obdélník 

   while (!q.empty()) {
      cur=q.front();
      q.pop();

      if (cur == cil) {
         i = 0;
         while (cur) {
            path[++i] = what[cur];
            cur = prev[cur];    
         }
         
         vystup = fopen("postup.out", "w");
         fprintf(vystup, "%d\n", i);
         while (i) {
            fprintf(vystup, "%c", path[i]);
            i--;
         }
         fclose(vystup);
         break;
      }
   
      dekoduj(cur, perm);

       
      // vygeneruj tři možné transformace aktuální permutace
      for (i = 0; i < 4; i++) {
         permn[0][i] = perm[7-i];
         permn[0][i+4] = perm[3-i];
         permn[1][i] = perm[(i+3)%4];
         permn[1][i+4] = perm[4+(i+1)%4];
      }
      permn[2][0] = perm[0];
      permn[2][1] = perm[6];
      permn[2][2] = perm[1];
      permn[2][3] = perm[3];
      permn[2][4] = perm[4];
      permn[2][5] = perm[2];
      permn[2][6] = perm[5];
      permn[2][7] = perm[7];
   
      for (i = 0; i < 3; i++) {
         int kod = zakoduj(permn[i]);
         if (!hotovo[kod]) {
            hotovo[kod] = 1;
            prev[kod] = cur;
            switch(i)
            {
            case 0:
                what[kod] = 'V';
                break;
            case 1:
                what[kod] = 'S';
                break;
            case 2:
                what[kod] = 'R';
                break;
            }
            q.push(kod);
         }  
      }
   }
   
   return 0;
}