Problem: 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.

Examples

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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long INF = 1000000000000000;
bool cmp(pair<string, long long>& a, pair<string, long long>& b) {
  return a.second < b.second;
}
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
long long factorial(long long n) {
  return (n == 1 || n == 0) ? 1 : (n * factorial(n - 1) % 1000000007);
}
string add(string a, string b) {
  if (a.size() < b.size()) swap(a, b);
  string ans;
  long long n = a.size();
  long long n1 = a.size() - b.size();
  string temp;
  for (long long i = 0; i < n1; i++) temp += "0";
  b = temp + b;
  long long car = 0;
  for (long long i = n - 1; i >= 0; i--) {
    ans += (char)((a[i] - '0' + b[i] - '0' + car) % 10 + '0');
    car = (a[i] - '0' + b[i] - '0' + car) / 10;
  }
  if (car) ans += (char)(car + '0');
  reverse(ans.begin(), ans.end());
  return ans;
}
bool cmp(string s1, string s2) {
  if (s1.size() != s2.size()) return s1.size() < s2.size();
  long long n = s1.size();
  for (long long i = 0; i < n; i++)
    if (s1[i] > s2[i])
      return 0;
    else if (s1[i] < s2[i])
      return 1;
  return 1;
}
vector<long long> z_function(string s) {
  long long n = (long long)s.length();
  vector<long long> z(n);
  for (long long i = 1, l = 0, r = 0; i < n; ++i) {
    if (i <= r) z[i] = min(r - i + 1, z[i - l]);
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
  }
  return z;
}
long long modp(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % 1000000007;
    b /= 2;
    a = (a * a) % 1000000007;
  }
  return ans;
}
void reverseStr(string& str) {
  long long n = str.length();
  for (long long i = 0; i < n / 2; i++) swap(str[i], str[n - i - 1]);
}
void bfs(vector<set<long long>> g, long long s, long long dist[]) {
  bool mark[1010];
  memset(mark, 0, sizeof mark);
  dist[s] = 0;
  queue<long long> q;
  q.push(s);
  mark[s] = 1;
  while (!q.empty()) {
    long long u = q.front();
    q.pop();
    for (auto v : g[u]) {
      if (mark[v]) continue;
      dist[v] = 1 + dist[u];
      mark[v] = 1;
      q.push(v);
    }
  }
}
vector<pair<long long, long long>> wg[100050];
long long f = 0;
vector<long long> d(100050, INF);
vector<long long> dijkstra(long long src) {
  set<pair<long long, long long>> s;
  d[src] = 0;
  s.insert({0, src});
  while (!s.empty()) {
    long long i = s.begin()->second;
    s.erase(s.begin());
    for (auto& e : wg[i]) {
      long long nb = e.first;
      long long w = e.second;
      if (d[i] + w < d[nb]) {
        s.erase({d[nb], nb});
        d[nb] = d[i] + w;
        s.insert({d[nb], nb});
      }
    }
  }
  return d;
}
long long lim = 150005;
long long sz;
vector<long long> adj[150005];
vector<long long> vis(lim, 0);
void dfs(long long x) {
  if (vis[x] == 1) return;
  sz++;
  vis[x] = 1;
  for (auto& it : adj[x]) {
    if (vis[it] != 1) {
      dfs(it);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  string s, t;
  cin >> s >> t;
  map<char, long long> cs;
  map<char, long long> ct;
  for (long long i = 0; i < s.length(); i++) {
    cs[s[i]]++;
  }
  for (long long i = 0; i < t.length(); i++) {
    ct[t[i]]++;
  }
  set<pair<long long, char>> need;
  for (char i = 'a'; i <= 'z'; i++) {
    if (ct[i] > 0) {
      need.insert({cs[i] / ct[i], i});
    }
  }
  for (long long i = 0; i < s.length(); i++) {
    if (s[i] == '?') {
      pair<long long, char> get = *(need.begin());
      need.erase(need.begin());
      long long c = get.first;
      char n = get.second;
      s[i] = n;
      cs[n]++;
      need.insert({cs[n] / ct[n], n});
    }
  }
  cout << s;
  return 0;
}