#include <bits/stdc++.h>
using namespace std;
template<class T, class U>
void ckmin(T &a, U b)
{
if (a > b) a = b;
}
template<class T, class U>
void ckmax(T &a, U b)
{
if (a < b) a = b;
}
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define SZ(x) ((int) ((x).size()))
#define ALL(x) (x).begin(), (x).end()
#define INF 1000000007
#define LLINF 2696969696969696969ll
#define MAXN 1100013
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
int N, Q;
int arr[MAXN];
int sos[MAXN], sob[MAXN];
vi freq[3];
int ans;
int32_t main()
{
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N >> Q;
string temps; cin >> temps;
FOR(i, 0, (1 << N))
{
arr[i] = temps[i] - '0';
sos[i] = arr[i];
sob[i] = arr[i];
}
//compute answer if there were only 1s and ????s
FOR(i, 0, N)
{
FOR(j, 0, (1 << N))
{
if (j & (1 << i))
{
sos[j] += sos[j - (1 << i)];
}
else
{
sob[j] += sob[j + (1 << i)];
}
}
}
while(Q--)
{
cin >> temps; reverse(ALL(temps));
ans = 0;
int tot = 0;
FOR(i, 0, 3) freq[i].clear();
FOR(i, 0, N)
{
int want = temps[i] - '0';
if (temps[i] == '?') want = 2;
freq[want].PB(i);
}
int mn = 0;
FOR(i, 0, 3) if (SZ(freq[i]) < SZ(freq[mn])) mn = i;
//now solve it in O(2^mn)
if (mn == 2)
{
int idx = 0;
for (int x : freq[1]) idx += (1 << x);
FOR(i, 0, (1 << SZ(freq[2])))
{
int cidx = idx;
FOR(j, 0, SZ(freq[2]))
{
if ((i & (1 << j)))
{
cidx += (1 << freq[2][j]);
}
}
ans += arr[cidx];
}
}
if (mn == 1)
{
int idx = 0;
for (int x : freq[1]) idx += (1 << x);
for (int x : freq[2]) idx += (1 << x);
FOR(i, 0, (1 << SZ(freq[1])))
{
int cidx = idx;
FOR(j, 0, SZ(freq[1]))
{
if ((i & (1 << j))) cidx -= (1 << freq[1][j]);
}
ans += ((__builtin_popcount(i) & 1) ? -1 : 1) * sos[cidx];
}
}
if (mn == 0)
{
int idx = 0;
for (int x : freq[1]) idx += (1 << x);
FOR(i, 0, (1 << SZ(freq[0])))
{
int cidx = idx;
FOR(j, 0, SZ(freq[0]))
{
if (i & (1 << j)) cidx += (1 << freq[0][j]);
}
ans += ((__builtin_popcount(i) & 1) ? -1 : 1) * sob[cidx];
}
}
cout << ans << '\n';
}
return 0;
}
