Problem:
Max Score: 250 Points

Problem Statement

Mr.X, who the handle name is T, looked at the list which written N handle names, S_1, S_2, ..., S_N.

But he couldn't see some parts of the list. Invisible part is denoted ?.

Please calculate all possible index of the handle name of Mr.X when you sort N+1 handle names (S_1, S_2, ..., S_N and T) in lexicographical order.

Note: If there are pair of people with same handle name, either one may come first.

Input

The input is given from standard input in the following format.

N
S_1
S_2
:
S_N
T

Output

Calculate the possible index and print in sorted order. The output should be separated with a space. Don't print a space after last number.

Put a line break in the end.

Constraints

1 ≤ N ≤ 10000

1 ≤ |S_i|, |T| ≤ 20 (|A| is the length of A.)

S_i consists from lower-case alphabet and ?.

T consists from lower-case alphabet.

Scoring

Subtask 1 [ 130 points ]

There are no ?'s.

Subtask 2 [ 120 points ]

There are no additional constraints.

Sample Input 1

2
tourist
petr
e

Sample Output 1

1

There are no invisible part, so there are only one possibility.
The sorted handle names are e, petr, tourist, so e comes first.

Sample Input 2

2
?o?r?s?
?et?
e

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <memory.h>
#include <math.h>
#include <assert.h>
#include <queue>
#include <map>
#include <set>
#include <string>
#include <algorithm>
#include <iostream>
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include <list>
#include <bitset>

using namespace std;
typedef pair<int, int> Pi;
typedef long long ll;
#define pii Pi
#define pll PL
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define sz(x) ((int)(x).size())
#define rep(i, n) for(int i=0;i<n;i++)
#define all(x) (x).begin(), (x).end()
typedef tuple<int, int, int> t3;
typedef tuple<int, int, int, int> t4;
typedef pair<ll, ll> PL;
typedef long double ldouble;

string A[10010];
string X;

void solve(){
	int n; cin >> n;
	rep(i, n)cin >> A[i];
	cin >> X;
	int l = 0, r = 0;
	rep(i, n){
		string p = A[i];
		for(auto &e : p)if(e == '?')e = 'a';
		if(p <= X)++r;
		p = A[i];
		for(auto &e : p)if(e == '?')e = 'z';
		if(p < X)++l;
	}
	for(int i=l;i<=r;i++)printf("%d ", i+1);
}

int main(){
	int Tc = 1; //scanf("%d\n", &Tc);
	for(int tc=1;tc<=Tc;tc++){
		//printf("Case #%d: ", tc);
		solve();
	}
	return 0;
}