Problem:
Score : 600 points

Problem Statement

There is a game that involves three variables, denoted A, B, and C.

As the game progresses, there will be N events where you are asked to make a choice.
Each of these choices is represented by a string s_i. If s_i is AB, you must add 1 to A or B then subtract 1 from the other; if s_i is AC, you must add 1 to A or C then subtract 1 from the other; if s_i is BC, you must add 1 to B or C then subtract 1 from the other.

After each choice, none of A, B, and C should be negative.

Determine whether it is possible to make N choices under this condition. If it is possible, also give one such way to make the choices.

Constraints

1 \leq N \leq 10^5

0 \leq A,B,C \leq 10^9

N, A, B, C are integers.

s_i is AB, AC, or BC.

Input

Input is given from Standard Input in the following format:

N A B C
s_1
s_2
:
s_N

Output

If it is possible to make N choices under the condition, print Yes; otherwise, print No.

Also, in the former case, show one such way to make the choices in the subsequent N lines. The (i+1)-th line should contain the name of the variable (A, B, or C) to which you add 1 in the i-th choice.

Sample Input 1

2 1 3 0
AB
AC

Sample Output 1

Yes
A
C

You can successfully make two choices, as follows:

In the first choice, add 1 to A and subtract 1 from B. A becomes 2, and B becomes 2.

In the second choice, add 1 to C and subtract 1 from A. C becomes 1, and A becomes 1.

Sample Input 2

3 1 0 0
AB
BC
AB

Sample Output 2

No

Sample Input 3

1 0 9 0
AC

Sample Output 3

No

Sample Input 4

8 6 9 1
AC
BC
AB
BC
AC
BC
AB
AB

Sample Output 4

Yes
C
B
B
C
C
B
A
A

Solution in Lua (LuaJIT 2.1.0) (judge verdict: Accepted)

local q, a, b, c = io.read("*n", "*n", "*n", "*n", "*l")
local op = {}
for i = 1, q do
  op[i] = io.read()
end
if a + b + c == 0 then
  print("No")
  os.exit()
elseif a + b + c == 1 then
  local ret = {}
  local valid = true
  for iq = 1, q do
    if op[iq] == "AB" then
      if a == 1 then
        ret[iq] = "B"
        a, b = 0, 1
      elseif b == 1 then
        ret[iq] = "A"
        a, b = 1, 0
      else
        valid = false break
      end
    elseif op[iq] == "BC" then
      if b == 1 then
        ret[iq] = "C"
        b, c = 0, 1
      elseif c == 1 then
        ret[iq] = "B"
        b, c = 1, 0
      else
        valid = false break
      end
    else
      if a == 1 then
        ret[iq] = "C"
        a, c = 0, 1
      elseif c == 1 then
        ret[iq] = "A"
        a, c = 1, 0
      else
        valid = false break
      end
    end
  end
  if valid then
    print("Yes")
    print(table.concat(ret, "\n"))
  else
    print("No")
  end
  os.exit()
elseif 3 <= a + b + c then
  if a + b == 0 and op[1] == "AB" then
    print("No") os.exit()
  elseif b + c == 0 and op[1] == "BC" then
    print("No") os.exit()
  elseif a + c == 0 and op[1] == "AC" then
    print("No") os.exit()
  end
  local ret = {}
  for iq = 1, q do
    if op[iq] == "AB" then
      if a < b then
        ret[iq] = "A"
        a, b = a + 1, b - 1
      else
        ret[iq] = "B"
        a, b = a - 1, b + 1
      end
    elseif op[iq] == "BC" then
      if b < c then
        ret[iq] = "B"
        b, c = b + 1, c - 1
      else
        ret[iq] = "C"
        b, c = b - 1, c + 1
      end
    else
      if a < c then
        ret[iq] = "A"
        a, c = a + 1, c - 1
      else
        ret[iq] = "C"
        a, c = a - 1, c + 1
      end
    end
  end
  print("Yes")
  print(table.concat(ret, "\n"))
  os.exit()
end
-- a + b + c == 2
local dp = {}
for j = 1, q + 1 do
  dp[j] = {}
  -- 110, 101, 011, 200, 020, 002
  for i = 1, 6 do
    dp[j][i] = false
  end
end
if a == 1 and b == 1 then
  dp[1][1] = true
elseif a == 1 and c == 1 then
  dp[1][2] = true
elseif b == 1 and c == 1 then
  dp[1][3] = true
elseif a == 2 then
  dp[1][4] = true
elseif b == 2 then
  dp[1][5] = true
else
  dp[1][6] = true
end
for i = 1, q do
  if op[i] == "AB" then
    if dp[i][1] then
      dp[i + 1][4] = true
      dp[i + 1][5] = true
    end
    if dp[i][2] then
      dp[i + 1][3] = true
    end
    if dp[i][3] then
      dp[i + 1][2] = true
    end
    if dp[i][4] or dp[i][5] then
      dp[i + 1][1] = true
    end
  elseif op[i] == "BC" then
    if dp[i][3] then
      dp[i + 1][5] = true
      dp[i + 1][6] = true
    end
    if dp[i][2] then
      dp[i + 1][1] = true
    end
    if dp[i][1] then
      dp[i + 1][2] = true
    end
    if dp[i][5] or dp[i][6] then
      dp[i + 1][3] = true
    end
  else
    if dp[i][2] then
      dp[i + 1][4] = true
      dp[i + 1][6] = true
    end
    if dp[i][1] then
      dp[i + 1][3] = true
    end
    if dp[i][3] then
      dp[i + 1][1] = true
    end
    if dp[i][4] or dp[i][6] then
      dp[i + 1][2] = true
    end
  end
end
local cur = false
for i = 1, 6 do
  if dp[q + 1][i] then
    cur = i
    break
  end
end
if not cur then
  print("No")
  os.exit()
end
local ret = {}
for i = 1, q do ret[i] = 0 end
for i = q, 1, -1 do
  local iq = i
  -- print(cur)
  if op[iq] == "AB" then
    if cur == 1 then
      if dp[i][4] then
        ret[i] = "B"
        cur = 4
      else--5
        ret[i] = "A"
        cur = 5
      end
    elseif cur == 2 then
      ret[i] = "A"
      cur = 3
    elseif cur == 3 then
      ret[i] = "B"
      cur = 2
    elseif cur == 4 then
      ret[i] = "A"
      cur = 1
    else-- 5
      ret[i] = "B"
      cur = 1
    end
  elseif op[iq] == "BC" then
    if cur == 3 then
      if dp[i][5] then
        ret[i] = "C"
        cur = 5
      else--6
        ret[i] = "B"
        cur = 6
      end
    elseif cur == 1 then
      ret[i] = "B"
      cur = 2
    elseif cur == 2 then
      ret[i] = "C"
      cur = 1
    elseif cur == 5 then
      ret[i] = "B"
      cur = 3
    else-- 6
      ret[i] = "C"
      cur = 3
    end
  else
    if cur == 2 then
      if dp[i][4] then
        ret[i] = "C"
        cur = 4
      else--6
        ret[i] = "A"
        cur = 6
      end
    elseif cur == 1 then
      ret[i] = "A"
      cur = 3
    elseif cur == 3 then
      ret[i] = "C"
      cur = 1
    elseif cur == 4 then
      ret[i] = "A"
      cur = 2
    else-- 6
      ret[i] = "C"
      cur = 2
    end
  end
end
print("Yes")
print(table.concat(ret, "\n"))