CodeForces

Codeforces Round #597 (Div. 2)

Let nn be a positive integer. Let a,b,ca,b,c be nonnegative integers such that a+b+c=na+b+c=n.

Alice and Bob are gonna play rock-paper-scissors nn times. Alice knows the sequences of hands that Bob will play. However, Alice has to play rock aa times, paper bb times, and scissors cc times.

Alice wins if she beats Bob in at least ⌈n2⌉⌈n2⌉ (n2n2 rounded up to the nearest integer) hands, otherwise Alice loses.

Note that in rock-paper-scissors:

• rock beats scissors;
• paper beats rock;
• scissors beat paper.

The task is, given the sequence of hands that Bob will play, and the numbers a,b,ca,b,c, determine whether or not Alice can win. And if so, find any possible sequence of hands that Alice can use to win.

If there are multiple answers, print any of them.

Input

The first line contains a single integer tt (1≤t≤1001≤t≤100) — the number of test cases.

Then, tt testcases follow, each consisting of three lines:

• The first line contains a single integer nn (1≤n≤1001≤n≤100).
• The second line contains three integers, a,b,ca,b,c (0≤a,b,c≤n0≤a,b,c≤n). It is guaranteed that a+b+c=na+b+c=n.
• The third line contains a string ss of length nn. ss is made up of only 'R', 'P', and 'S'. The ii-th character is 'R' if for his ii-th Bob plays rock, 'P' if paper, and 'S' if scissors.

Output

For each testcase:

• If Alice cannot win, print "NO" (without the quotes).
• Otherwise, print "YES" (without the quotes). Also, print a string tt of length nn made up of only 'R', 'P', and 'S' — a sequence of hands that Alice can use to win. tt must contain exactly aa 'R's, bb 'P's, and cc 'S's.
• If there are multiple answers, print any of them.

The "YES" / "NO" part of the output is case-insensitive (i.e. "yEs", "no" or "YEs" are all valid answers). Note that 'R', 'P' and 'S' are case-sensitive.

Example

Input

2
3
1 1 1
RPS
3
3 0 0
RPS

Output

YES
PSR
NO

Note

In the first testcase, in the first hand, Alice plays paper and Bob plays rock, so Alice beats Bob. In the second hand, Alice plays scissors and Bob plays paper, so Alice beats Bob. In the third hand, Alice plays rock and Bob plays scissors, so Alice beats Bob. Alice beat Bob 3 times, and 3≥⌈32⌉=23≥⌈32⌉=2, so Alice wins.

In the second testcase, the only sequence of hands that Alice can play is "RRR". Alice beats Bob only in the last hand, so Alice can't win. 1<⌈32⌉=21<⌈32⌉=2.

水题，就是尽量赢，等不能赢了，剩下的看还能出什么，随便输出就行。

#include<bits/stdc++.h>
using namespace std;
char W[110000];
string s;
int main()
{
    int t;
    cin >> t;
    while(t--)
    {
        int R, P, S,n;
        cin >> n;
        cin >> R >> P >> S;
        cin >> s;
        int sum = 0;
        for (int i = 0;i < n;i++)
        {
            if (s[i] == 'R' && P > 0)
                W[i] = 'P', P--;
            else if (s[i] == 'P' && S > 0)
                W[i] = 'S', S--;
            else if (s[i] == 'S' && R > 0)
                W[i] = 'R', R--;
            else
                W[i] = 'N', sum++;
        }
        if (sum > n/2)
        {
            puts("NO");
            continue;
        }
         puts("YES");
        for (int i = 0;i < n;i++)
        {
            if (W[i] == 'N')
            {
                if (R > 0)
                {
                    cout << 'R';
                    R--;
                }
                else if (P > 0)
                {
                    cout << 'P';
                    P--;
                }
                else if (S > 0)
                {
                    cout << 'S';
                    S--;
                }
            }
            else
                cout << W[i];
        }
       puts("");

    }

    return 0;
}