炒鸡大水题！
从高到低枚举每一位能不能取1，然后分别算出该位取0和取1的答案再比对。
// It is made by XZZ
#include<cstdio>
#include<algorithm>
#define il inline
#define rg register
#define vd void
#define sta static
typedef long long ll;
il int gi(){
rg int x=0,f=1;rg char ch=getchar();
while(ch<'0'||ch>'9')f=ch=='-'?-1:f,ch=getchar();
while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
return x*f;
}
const int maxn=1e5+1;
int a[maxn],op[maxn];
char opt[10];
il int calc(int x,int y,int o){
if(o==1)return x&y;
else if(o==2)return x|y;
else return x^y;
}
int main(){
int n=gi(),m=gi();
for(rg int i=1;i<=n;++i){
scanf("%s",opt);a[i]=gi();
if(opt[0]=='A')op[i]=1;
else if(opt[0]=='O')op[i]=2;
else op[i]=3;
}
int A0=0,A1=(1<<30)-1;
for(rg int j=1;j<=n;++j)A0=calc(A0,a[j],op[j]);
for(rg int j=1;j<=n;++j)A1=calc(A1,a[j],op[j]);
int ans=0,atk=0;
bool x,a0,a1;
for(rg int i=29;~i;--i){
a0=A0&(1<<i);
a1=A1&(1<<i);
if(atk+(1<<i)>m)a1=0;
if(a1==1&&a0==0)atk+=1<<i,ans+=1<<i;
else if(a0==1)ans+=1<<i;
}
printf("%d
",ans);
return 0;
}