Problem: Doctor prescribed medicine to his patient. The medicine is represented by pills. Each pill consists of a shell and healing powder. The shell consists of two halves; each half has one of four colors — blue, red, white or yellow.

The doctor wants to put 28 pills in a rectangular box 7 × 8 in size. Besides, each pill occupies exactly two neighboring cells and any cell contains exactly one half of a pill. Thus, the result is a four colored picture 7 × 8 in size.

The doctor thinks that a patient will recover sooner if the picture made by the pills will be special. Unfortunately, putting the pills in the box so as to get the required picture is not a very easy task. That's why doctor asks you to help. 

Doctor has some amount of pills of each of 10 painting types. They all contain the same medicine, that's why it doesn't matter which 28 of them will be stored inside the box.

Place the pills in the box so that the required picture was formed. If it is impossible to place the pills in the required manner, then place them so that the number of matching colors in all 56 cells in the final arrangement and the doctor's picture were maximum.

Input

First 7 lines contain the doctor's picture. Each line contains 8 characters, each character can be "B", "R", "W" and "Y" that stands for blue, red, white and yellow colors correspondingly.

Next four lines contain 10 numbers that stand for, correspondingly, the number of pills painted:

"BY" "BW" "BR" "BB"

"RY" "RW" "RR"

"WY" "WW"

"YY"

Those numbers lie within range from 0 to 28 inclusively. It is guaranteed that the total number of pills in no less than 28.

Output

Print on the first line the maximal number cells for which the colors match.

Then print 13 lines each containing 15 characters — the pills' position in the optimal arrangement. The intersections of odd lines and odd columns should contain characters "B", "R", "W" and "Y". All other positions should contain characters ".", "-" and "|". Use "-" and "|" to show which halves belong to one pill. See the samples for more clarification.

If there are several possible solutions, print any of them.

Examples

Input

WWWBBWWW
WWWBBWWW
YYWBBWWW
YYWBBWRR
YYWBBWRR
YYWBBWRR
YYWBBWRR
0 0 0 8
0 1 5
1 10
5


Output

53
W.W.W.B.B.W.W.W
|.|.|.|.|.|.|.|
W.W.W.B.B.W.W.W
...............
Y.Y.W.B.B.W.W-W
|.|.|.|.|.|....
Y.Y.W.B.B.W.R.R
............|.|
Y.Y.W.B.B.R.R.R
|.|.|.|.|.|....
Y.Y.W.B.B.W.R.R
............|.|
Y-Y.B-B.B-B.R.R


Input

WWWWWWWW
WBYWRBBY
BRYRBWYY
WWBRYWBB
BWWRWBYW
RBWRBWYY
WWWWWWWW
0 0 0 1
0 0 1
0 1
25


Output

15
W.Y.Y-Y.Y-Y.Y-Y
|.|............
W.Y.Y.Y.Y.B-B.Y
....|.|.|.....|
Y-Y.Y.Y.Y.Y-Y.Y
...............
Y.Y.Y.R.Y.Y.Y-Y
|.|.|.|.|.|....
Y.Y.Y.R.Y.Y.Y.Y
............|.|
Y-Y.Y.Y-Y.Y.Y.Y
....|.....|....
Y-Y.Y.Y-Y.Y.Y-Y

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
class Node {
 public:
  int r, c;
  Node *prev, *next;
  Node() {}
  Node(int ri, int ci, Node *p, Node *n) : r(ri), c(ci), prev(p), next(n) {}
  void remove() {
    prev->next = next;
    next->prev = prev;
  }
  void recover() { prev->next = next->prev = this; }
};
const int inf = 1000000000;
const int mat[2][2] = {1, 0, 0, 1};
char c2d[128], d2c[4] = {'B', 'R', 'W', 'Y'};
const int rn = 7;
const int cn = 8;
char gr[rn][cn + 1];
int pill[10], pid[4][4], idp[10][2];
int fail[10] = {0}, failind[28];
int lsum, rsum;
const int src = 0, sink = 21, vn = 22;
int res[vn][vn] = {0}, initr[vn][vn];
int fv[vn] = {0}, fvid = 0;
Node head, tail, nd[56];
bool vis[rn][cn] = {0};
pair<int, int> con[28][2];
char aa[rn][cn + 1];
vector<int> with[10], none;
set<long long> vvv;
int opt;
pair<int, int> optc[28][2];
int typen[10] = {0};
const int frn = rn * 2 - 1, fcn = cn * 2 - 1;
char fig[frn][fcn + 1];
inline int id(int r, int c) { return r * cn + c; }
inline bool valid(int r, int c) { return r >= 0 && r < rn && c >= 0 && c < cn; }
inline int left(int x) { return x + 1; }
inline int right(int x) { return x + 11; }
inline void pre() {
  for (int i = 0; i < 10; i++) {
    int i0 = idp[i][0];
    int i1 = idp[i][1];
    for (int j = 0; j < 10; j++) {
      if (i == j) continue;
      int j0 = idp[j][0];
      int j1 = idp[j][1];
      if (i != j && (i0 == j0 || i0 == j1 || i1 == j0 || i1 == j1))
        initr[i][j] = inf;
      else
        initr[i][j] = 0;
    }
  }
}
inline void construct() {
  for (int i = 0; i < 10; i++) {
    for (int j = 0; j < 10; j++) {
      res[left(i)][right(j)] = initr[i][j];
      res[right(j)][left(i)] = 0;
    }
    res[right(i)][sink] = pill[i];
    res[sink][right(i)] = 0;
    res[src][left(i)] = fail[i];
    res[left(i)][src] = 0;
  }
}
inline int min(int a, int b) { return a < b ? a : b; }
inline int min(int a, int b, int c) {
  if (a < b) {
    if (a < c)
      return a;
    else
      return c;
  } else {
    if (b < c)
      return b;
    else
      return c;
  }
}
int fdfs(int v, int f) {
  fv[v] = fvid;
  if (res[v][sink]) {
    int a = min(f, res[v][sink]);
    res[v][sink] -= a;
    res[sink][v] += a;
    return a;
  }
  int u0 = (v > 10 || v == src ? 1 : 11), u1 = (v > 10 || v == src ? 10 : 20);
  for (int u = u0; u <= u1; u++) {
    if (fv[u] == fvid || !res[v][u]) continue;
    int a = fdfs(u, min(f, res[v][u]));
    if (a) {
      res[v][u] -= a;
      res[u][v] += a;
      return a;
    }
  }
  return 0;
}
inline int asdf() {
  int f = 0;
  for (int i = 0; i < 10; i++) {
    int l = left(i);
    if (!res[src][l]) continue;
    for (int j = 0; j < 10; j++) {
      int r = right(j);
      int a = min(res[src][l], res[l][r], res[r][sink]);
      if (!a) continue;
      f += a;
      res[r][sink] -= a;
      res[l][r] -= a;
      res[src][l] -= a;
      res[sink][r] += a;
      res[r][l] += a;
      res[l][src] += a;
    }
  }
  return f;
}
inline int flow() {
  int f = 0, a;
  f += asdf();
  while (1) {
    ++fvid;
    a = fdfs(src, inf);
    if (!a) return f;
    f += a;
  }
}
inline long long rep() {
  long long s = 0;
  for (int i = 0; i < 10; i++) s = s * 28 + typen[i];
  return s;
}
void dfs(Node *pt, int p, int cur) {
  if (cur + rsum + (28 - p) <= opt) return;
  if (pt == &tail) {
    if (cur + min(lsum, rsum) <= opt) return;
    long long rr = rep();
    if (vvv.find(rr) != vvv.end()) return;
    vvv.insert(rr);
    int f;
    construct();
    f = flow();
    if (cur + f > opt) {
      opt = cur + f;
      for (int i = 0; i < 28; i++)
        for (int j = 0; j < 2; j++) optc[i][j] = con[i][j];
      for (int i = 0; i < 28; i++) {
        int r1 = optc[i][0].first, c1 = optc[i][0].second;
        int r2 = optc[i][1].first, c2 = optc[i][1].second;
        int x1 = gr[r1][c1], x2 = gr[r2][c2];
        int pi = pid[x1][x2];
        if (r1 == r2)
          fig[r1 * 2][c1 * 2 + 1] = '-';
        else
          fig[r1 * 2 + 1][c1 * 2] = '|';
      }
      for (int i = 0; i < rn; i++)
        for (int j = 0; j < cn; j++) aa[i][j] = gr[i][j];
      none.clear();
      for (int l = 0; l < 10; l++) {
        with[l].clear();
        for (int r = 0; r < 10; r++) {
          int t = res[right(r)][left(l)];
          for (int i = 0; i < t; i++) with[l].push_back(r);
          t = res[right(r)][sink];
          for (int i = 0; i < t; i++) none.push_back(r);
        }
      }
      for (int i = 0; i < lsum; i++) {
        int r1 = optc[failind[i]][0].first, c1 = optc[failind[i]][0].second;
        int r2 = optc[failind[i]][1].first, c2 = optc[failind[i]][1].second;
        int x1 = gr[r1][c1], x2 = gr[r2][c2];
        int pi = pid[x1][x2];
        if (with[pi].size()) {
          int y1 = idp[with[pi].back()][0], y2 = idp[with[pi].back()][1];
          with[pi].pop_back();
          if (x1 == y1)
            aa[r2][c2] = y2;
          else if (x1 == y2)
            aa[r2][c2] = y1;
          else if (x2 == y1)
            aa[r1][c1] = y2;
          else if (x2 == y2)
            aa[r1][c1] = y1;
        } else {
          int y1 = idp[none.back()][0], y2 = idp[none.back()][1];
          none.pop_back();
          aa[r1][c1] = y1;
          aa[r2][c2] = y2;
        }
      }
    }
    return;
  }
  int r = pt->r, c = pt->c;
  int cx = gr[r][c];
  for (int i = 0; i < 2; i++) {
    int nr = r + mat[i][0];
    int nc = c + mat[i][1];
    if (!valid(nr, nc) || vis[nr][nc]) continue;
    int nx = gr[nr][nc];
    int pi = pid[cx][nx];
    bool has = pill[pi] ? 1 : 0;
    vis[nr][nc] = 1;
    nd[id(nr, nc)].remove();
    if (has)
      pill[pi]--, rsum--;
    else {
      failind[lsum] = p;
      fail[pi]++, lsum++;
    }
    con[p][0] = make_pair(r, c);
    con[p][1] = make_pair(nr, nc);
    typen[pi]++;
    dfs(pt->next, p + 1, cur + (has ? 2 : 0));
    typen[pi]--;
    if (has)
      pill[pi]++, rsum++;
    else
      fail[pi]--, lsum--;
    nd[id(nr, nc)].recover();
    vis[nr][nc] = 0;
  }
}
inline void output() {
  printf("%d\n", opt);
  memset(fig, '.', sizeof(fig));
  for (int i = 0; i < rn; i++)
    for (int j = 0; j < cn; j++) fig[i * 2][j * 2] = d2c[aa[i][j]];
  for (int i = 0; i < 28; i++) {
    int r1 = optc[i][0].first, c1 = optc[i][0].second;
    int r2 = optc[i][1].first, c2 = optc[i][1].second;
    if (r1 == r2)
      fig[r1 * 2][c1 * 2 + 1] = '-';
    else
      fig[r1 * 2 + 1][c1 * 2] = '|';
  }
  for (int i = 0; i < frn; i++) {
    fig[i][fcn] = '\0';
    puts(fig[i]);
  }
}
inline void solve() {
  head = Node(-1, -1, NULL, nd);
  tail = Node(-1, -1, nd + 55, NULL);
  for (int i = 0; i < rn; i++)
    for (int j = 0; j < cn; j++)
      nd[id(i, j)] = Node(i, j, id(i, j) ? nd + id(i, j) - 1 : &head,
                          id(i, j) < 55 ? nd + id(i, j) + 1 : &tail);
  opt = -1;
  dfs(nd + id(0, 0), 0, 0);
  output();
}
int main(void) {
  c2d['B'] = 0;
  c2d['R'] = 1;
  c2d['W'] = 2;
  c2d['Y'] = 3;
  for (int i = 0; i < rn; i++) {
    scanf("%s", gr[i]);
    for (int j = 0; j < cn; j++) gr[i][j] = c2d[gr[i][j]];
  }
  int pi = 0;
  lsum = rsum = 0;
  for (int i = 0; i < 4; i++) {
    for (int j = 3; j >= i; j--) {
      scanf("%d", pill + pi);
      rsum += pill[pi];
      idp[pi][0] = i;
      idp[pi][1] = j;
      pid[i][j] = pid[j][i] = pi++;
    }
  }
  pre();
  solve();
  return 0;
}