#include <bits/stdc++.h>
using namespace std;
char buf[1<<20], obuf[1<<20], *p1=buf, *p2=buf, *p3=obuf;
#define getchar() (p1==p2&&((p2=(p1=buf)+fread(buf,1,1<<20,stdin))&&p1==p2)?EOF:*p1++)
#define putchar(x) (p3-obuf<(1<<20)?*p3++=x:(fwrite(obuf,p3-obuf,1,stdout),p3=obuf,*p3++=x))
constexpr char space = '\n';
template<typename T> inline void read(T& x)
{
    char c=getchar(); x=0; bool k=1;
    for (;c<'0'||c>'9';c=getchar()) if (c=='-') k=0;
    for (;c>='0'&&c<='9';c=getchar()) x=x*10+(c^48);
    if (!k) x=-x;
}
inline char readc()
{
    char c=getchar();
    while (c==' '||c=='\n'||c=='\r') c=getchar();
    return c;
}
template <typename T, typename... arg>
inline void read(T& x, arg&... args) {read(x); read(args...);}
using i64 = long long;
using u32 = unsigned;
using u64 = unsigned long long;
constexpr i64 inf = 0x3f3f3f3f3f3f3f3f;
u64 o[4][256 + 3][1000000 / 64 + 5];
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, q;
    read(n, q);
    int t = 0;
    for (int i=0;i<q;++i)
    {
        char p = readc();
        u32 x;
        read(x);
        u32 a = x & 255, b = x >> 8 & 255, c = x >> 16 & 255, d = x >> 24 & 255;
        if (p == '?')
        {
            u64 e0 = 0, e1 = 0, e2 = 0, e3 = 0;
            int p = t >> 6, j = 0;
            for (; j<=p-3; j+=4)
            {
                e0 ^= o[0][a][j + 0] & o[1][b][j + 0] & o[2][c][j + 0] & o[3][d][j + 0];
                e1 ^= o[0][a][j + 1] & o[1][b][j + 1] & o[2][c][j + 1] & o[3][d][j + 1];
                e2 ^= o[0][a][j + 2] & o[1][b][j + 2] & o[2][c][j + 2] & o[3][d][j + 2];
                e3 ^= o[0][a][j + 3] & o[1][b][j + 3] & o[2][c][j + 3] & o[3][d][j + 3];
            }
            u64 e = e0 ^ e1 ^ e2 ^ e3;
            for (; j<=p; ++j) e ^= o[0][a][j] & o[1][b][j] & o[2][c][j] & o[3][d][j];
            putchar(__builtin_parityll(e) + '0');
            putchar('\n');
        }
        else
        {
            u64 v = 1ull << (t & 63);
            u32 p = t >> 6;
            ++t;
            for (int j = a; j < 256; j = (j + 1) | a) o[0][j][p] |= v;
            for (int j = b; j < 256; j = (j + 1) | b) o[1][j][p] |= v;
            for (int j = c; j < 256; j = (j + 1) | c) o[2][j][p] |= v;
            for (int j = d; j < 256; j = (j + 1) | d) o[3][j][p] |= v;
        }
    }
    fwrite(obuf, p3 - obuf, 1, stdout);
    return 0;
}