// Problem: 起床困难综合症
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1053
// Memory Limit: 1048576 MB
// Time Limit: 2000 ms
// 2022-02-22 18:59:43
//
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline int read () {
int x = 0, f = 0;
char ch = getchar();
while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
return f?-x:x;
}
template<typename T> void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x >= 10) print(x/10);
putchar(x % 10 + '0');
}
ll n, m;
void solve() {
bitset<40> zero, one;
zero.reset();
one.set();
cin >> n >> m;
for (int i = 0; i < n; i ++) {
string s;
cin >> s;
int x;
cin >> x;
if (s[0] == 'A') zero &= x, one &= x;
if (s[0] =='O') zero |= x, one |= x;
if (s[0] == 'X') zero ^= x, one ^= x;
}
ll ans = 0;
for (int i = 30; i >= 0; i --)
{
ll t = 1 << i;
if (zero[i]) ans += t;
else if (one[i] && m - t >= 0) ans += t;
}
cout << ans << endl;
}
int main () {
int t;
t =1;
//cin >> t;
while (t --) solve();
return 0;
}


从每一位进行思考。对于最终的结果。如果为1,那么选择数在这位上可能是1或者是0.如果是0那么必选,否则要比较和m的大小再考虑选不选 我是知道在位上进行思考。但是是从每次选择某个数的条件下的角度进行想的。