Problem: For strings s and t, we will say that s and t are prefix-free when neither is a prefix of the other.

Let L be a positive integer. A set of strings S is a good string set when the following conditions hold true:

* Each string in S has a length between 1 and L (inclusive) and consists of the characters `0` and `1`.
* Any two distinct strings in S are prefix-free.



We have a good string set S = \\{ s_1, s_2, ..., s_N \\}. Alice and Bob will play a game against each other. They will alternately perform the following operation, starting from Alice:

* Add a new string to S. After addition, S must still be a good string set.



The first player who becomes unable to perform the operation loses the game. Determine the winner of the game when both players play optimally.

Constraints

* 1 \leq N \leq 10^5
* 1 \leq L \leq 10^{18}
* s_1, s_2, ..., s_N are all distinct.
* { s_1, s_2, ..., s_N } is a good string set.
* |s_1| + |s_2| + ... + |s_N| \leq 10^5

Input

Input is given from Standard Input in the following format:


N L
s_1
s_2
:
s_N


Output

If Alice will win, print `Alice`; if Bob will win, print `Bob`.

Examples

Input

2 2
00
01


Output

Alice


Input

2 2
00
11


Output

Bob


Input

3 3
0
10
110


Output

Alice


Input

2 1
0
1


Output

Bob


Input

1 2
11


Output

Alice


Input

2 3
101
11


Output

Bob

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#define be(v) (v).begin(),(v).end()
#define pb(q) push_back(q)
typedef long long ll;
using namespace std;
const ll mod=1000000007, INF=mod*mod*3LL;
#define doublecout(a) cout<<fixed<<setprecision(10)<<a<<endl;
ll n ,L;
map<ll,bool> mp;
vector<string> s;
void dfs(ll len, ll le, ll ri){
	if(ri - le == 0){
		mp[L - len + 1] ^= 1;
		return;
	}
	if(ri - le == 1){
		if(len == s[le].length()) return;
		dfs(len + 1, le, ri);
		dfs(len + 1, le + 1, ri);
		return;
	}
	ll mi = le - 1, ma = ri, mid;
	while(ma - mi > 1){
		mid = (ma + mi) / 2;
		if(s[mid][len]=='0') mi = mid;
		else ma = mid;
	}
	dfs(len + 1, le, mi + 1);
	dfs(len + 1, ma, ri);
	return;
}
int main() {
    cin.tie(0);
    cout.tie(0);
    ios::sync_with_stdio(false);
    cin >> n >> L;
    s.resize(n);
    for(int i=0;i<n;i++){
    	cin >> s[i];
    }
    sort(be(s));
    dfs(0, 0, n);
    bool maki = false;
    map<ll,bool> g;
    for(auto& i:mp){
    	g[i.first&(-i.first)] ^= i.second;
    }
    for(auto& i:g){
    	if(i.second) maki = true;
    }
    cout << (maki?"Alice":"Bob") << endl;
    return 0;
}