描述

You are given two strings s and t consisting of small Latin letters, string s can also contain ‘?’ characters.

Suitability of string s is calculated by following metric:

Any two letters can be swapped positions, these operations can be performed arbitrary number of times over any pair of positions. Among all resulting strings s

, you choose the one with the largest number of non-intersecting occurrences of string t. Suitability is this number of occurrences.

You should replace all ‘?’ characters with small Latin letters in such a way that the suitability of string s is maximal.

#Input

The first line contains string s (1 ≤ |s

| ≤ 106).

The second line contains string t (1 ≤ |t| ≤ 106).

#Output

Print string s with ‘?’ replaced with small Latin letters in such a way that suitability of that string is maximal.

If there are multiple strings with maximal suitability then print any of them.

input

?aa?
ab

output

baab

input

??b?
za
 

output

azbz

input

abcd
abacaba

output

abcd

Note

In the first example string “baab” can be transformed to “abab” with swaps, this one has suitability of 2. That means that string “baab” also has suitability of 2.

In the second example maximal suitability you can achieve is 1 and there are several dozens of such strings, “azbz” is just one of them.

In the third example there are no ‘?’ characters and the suitability of the string is 0.

思路

給你兩個串s和t，s串中可能包含若干個?，現在已知s串中的字母可以任意交換位置，題目讓你把?填滿，使得最終的s串包含的t串最多。

我們可以用vextor存一下?的位置，然後貪心的不斷的列舉t串中的字元，如果字元在a中存在則不填，否則填，直到把問號填完。

程式碼

#include <bits/stdc++.h>
using namespace std;
int a[300];
int main()
{
    vector<int> pos;
    string s, t;
    cin >> s >> t;
    for (int i = 0; i < s.size(); i++)
    {
        if (s[i] == '?')
            pos.push_back(i);
        a[s[i]]++;
    }
    int flag = 1;
    while (flag)
        for (char ch : t)
            if (a[ch])
                a[ch]--;
            else if (pos.size())
            {
                s[pos[pos.size() - 1]] = ch;
                pos.pop_back();
            }
            else
            {
                flag = 0;
                break;
            }
    cout << s << endl;

    return 0;
}