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 C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#pragma GCC optimize("O2")
using namespace std;
#define FASTIO ios_base::sync_with_stdio(false), cin.tie(0)
//#define MULTI_TEST
//#define GOOGLE
#ifdef LOCAL
#include "debugger.h"
#else
#define db(...)
#endif
#define ll long long
#define v32 vector<int>
#define v64 vector<ll>
#define s32 set<int>
#define s64 set<ll>
#define p32 pair<int, int>
#define p64 pair<ll, ll>
#define sz(v) (ll)(v).size()
#define fi first
#define se second
#define ln '\n'
const ll MOD = 1e9 + 7;
inline ll add(ll a, ll b, ll m) { return (((a % m + b % m) % m + m) % m); }
inline ll mul(ll a, ll b, ll m) { return (((a % m * b % m) % m + m) % m); }

void solve()
{
	ll n, a, b, c;
	cin >> n >> a >> b >> c;
	v32 curr(3, 0);
	curr[0] = a, curr[1] = b, curr[2] = c;
	vector<string> arr(n);
	for (int i = 0; i < n; i++)
	{
		cin >> arr[i];
	}
	vector<char> ans;
	for (int i = 0; i < n; i++)
	{
		int idx1 = arr[i][0] - 'A', idx2 = arr[i][1] - 'A';
		if (curr[idx1] < curr[idx2])
		{
			curr[idx1]++;
			curr[idx2]--;
			ans.push_back(arr[i][0]);
		}
		else if (curr[idx1] > curr[idx2])
		{
			curr[idx1]--;
			curr[idx2]++;
			ans.push_back(arr[i][1]);
		}
		else
		{
			int next_idx1 = -1, next_idx2 = -1;
			if (i + 1 < n)
			{
				if (arr[i + 1] == arr[i])
				{
					curr[idx1]--;
					curr[idx2]++;
					ans.push_back(arr[i][1]);
					continue;
				}
				next_idx1 = arr[i + 1][0] - 'A';
				next_idx2 = arr[i + 1][1] - 'A';
			}
			if (idx1 > idx2)
				swap(idx1, idx2);
			if (next_idx1 > next_idx2)
				swap(next_idx1, next_idx2);
			if (idx1 == next_idx1)
			{
				curr[idx1]++;
				curr[idx2]--;
				ans.push_back(idx1 + 'A');
			}
			else
			{
				curr[idx1]--;
				curr[idx2]++;
				ans.push_back(idx2 + 'A');
			}
		}
	}
	for (int x : curr)
	{
		if (x < 0)
		{
			cout << "No" << ln;
			return;
		}
	}
	cout << "Yes" << ln;
	for (char c : ans)
		cout << c << ln;
	return;
}

int main()
{
#ifndef LOCAL
	FASTIO;
#endif
	int t = 1;
#ifdef LOCAL
	clock_t start, stop;
	start = clock();
#endif
#ifdef MULTI_TEST
	cin >> t;
#endif
	for (int tc = 1; tc < t + 1; tc++)
	{
#ifdef GOOGLE
		cout << "Case #" << tc << ": ";
#endif
		solve();
	}
#ifdef LOCAL
	stop = clock();
	cerr << "Time :" << fixed << setprecision(5) << double(stop - start) / double(CLOCKS_PER_SEC) << "s" << ln;
#endif
	return 0;
}