stablegradients/pie-gem5-pairs huggingface dataset:
// Slower C++ solution:
#include<bits/stdc++.h>

#define MAXN 100005

#define reg register

#define inl inline

using namespace std;

char s[MAXN],ch[MAXN];

inl char Chg(reg char x)

{

	return x=='b'?'d':(x=='d'?'b':(x=='p'?'q':'p'));

}

int main()

{

	scanf("%s",ch+1);

	reg int len=strlen(ch+1);

	memcpy(s,ch,sizeof(ch));

	reverse(ch+1,ch+len+1);

	for(reg int i=1;i<=len;i++) ch[i]=Chg(ch[i]);

	reg bool fg=1;

	for(reg int i=1;i<=len;i++)

	{

		if(ch[i]!=s[i])

		{

			fg=0;

			break;

		}

	}

	puts(fg?"Yes":"No");

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1919475380933966x):
#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<cstdlib>

#include<algorithm>

#include<string>

#include<queue>

#define re register

#define ll long long 

#define maxn 10010000

#define mod 6662333 

#define inf 2147483647

#define mem(p) memset(&p,0,sizeof(p))

using namespace std;

inline int read()

{ 

    int x=0,f=1;char c=getchar();

    while(!isdigit(c)){if(c=='-')f=-1;c=getchar();} 

    while(isdigit(c)){x=x*10+c-48;c=getchar();}

    return x*f;

}

inline void put(ll x)

{

     if(x<0) putchar('-'),x=-x;

     if(x>9) put(x/10);

     putchar(x%10+'0');

}

char a[maxn],b[maxn];

int sum;

int main()

{

    cin>>a;

    sum=strlen(a);//求字符数组a的长度 

    sum-=1;

    for(int i=0;a[i]!='\0';i++)

	{

        if(a[i]=='b') b[sum]='d';

        if(a[i]=='d') b[sum]='b';

        if(a[i]=='p') b[sum]='q';

        if(a[i]=='q') b[sum]='p';

        sum--;

    }

    if(strcmp(a,b)==false) cout<<"Yes";

	else cout<<"No";

    return 0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define MAXN 100005

#define reg register

#define inl inline

using namespace std;

char s[MAXN],ch[MAXN];

inl char Chg(reg char x)

{

	return x=='b'?'d':(x=='d'?'b':(x=='p'?'q':'p'));

}

int main()

{

	scanf("%s",ch+1);

	reg int len=strlen(ch+1);

	memcpy(s,ch,sizeof(ch));

	reverse(ch+1,ch+len+1);

	for(reg int i=1;i<=len;i++) ch[i]=Chg(ch[i]);

	reg bool fg=1;

	for(reg int i=1;i<=len;i++)

	{

		if(ch[i]!=s[i])

		{

			fg=0;

			break;

		}

	}

	puts(fg?"Yes":"No");

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 1.1887026508343241x):
#include<iostream>

#include<string> 

using namespace std;

string s,s2;

int main()

{

	cin>>s;

	for(int i=s.size()-1;i>=0;i--)

	{

		if(s[i]=='b') s2+='d';

		else if(s[i]=='d') s2+='b';

		else if(s[i]=='q') s2+='p';

		else s2+='q';

	}

	if(s2==s) cout<<"Yes"<<endl;

	else cout<<"No"<<endl;

	return 0;

}

/*



*/
// Slower C++ solution:
#include<iostream>

#include<cstdio>

#include<cstring>

#include<cmath>;

using namespace std;

char s[100010];

int main(){

    cin>>s;

    int k=strlen(s);

    if(k%2!=0){

        printf("No");

        return 0;

    }

    for(int i=0;i<k/2;i++){

        int u=abs(s[i]-s[k-i-1]);

        if(u!=1&&u!=2){

            printf("No");

            return 0;

        }

    }

    printf("Yes");

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.30710880182067x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

bool is;

char ch,now,a;

string s;

stack<int> t;

int main()

{

	cin>>s; int n=s.size();

	for(int i=0;i<n;i++)

	    t.push(s[i]);

	for(int i=0;i<n;i++)

	{

		now=t.top();

	    if(now=='b') a='d';

	    else if(now=='d') a='b';

	    else if(now=='p') a='q';

	    else if(now=='q') a='p';

	    if(a==s[i]) t.pop();

		else

		{

			is=1; break;

		}

	}

	if(!is) cout<<"Yes"<<endl;

	else cout<<"No"<<endl;

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.383965163079413x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{

	string s;

	cin>>s;

	if(s.size()&1) 

	{

        printf("No");

        return 0;

    }

	for(int i=0; i<s.size()/2; i++) 

	{

        int r=abs(s[i]-s[s.size()-i-1]);

        if(r!=1&&r!=2) 

        { 

            cout<<"No";

            return 0;

        }

    }

	cout<<"Yes";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.357776416211624x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

#define l strlen(a)

using namespace std;

char a[100000];int x;

int main()

{

	scanf("%s",&a);x=abs(a[0]-a[l-1]);

	if(l%2==1){cout<<"No\n";return 0;}

	for(int i=0;i<l/2;i++,x=abs(a[i]-a[l-i-1]))

	  if(x!=1&&x!=2){cout<<"No\n";return 0;}

	cout<<"Yes\n";

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.279253359671471x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{char a[100100],b[100100];

 cin>>a;

 strcpy(b,a);

 for(int i=0;i<strlen(b);i++)

    {if(b[i]=='b') b[i]='d';

     else if(b[i]=='p') b[i]='q';

     else if(b[i]=='q') b[i]='p';

     else if(b[i]=='d') b[i]='b';

	}

 for(int i=0;i<strlen(a);i++)

    if(a[i]!=b[strlen(a)-i-1])

      {printf("No\n");

       return 0;

	  }

 printf("Yes\n");

 return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.310615287879859x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

typedef long long LL;

using namespace std;



namespace IO {

    inline LL read() {

        LL o = 0, f = 1; char c = getchar();

        while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }

        while (c > '/' && c < ':') { o = o * 10 + c - '0'; c = getchar(); }

        return o * f;

    }

    inline char recd() {

        char o; while ((o = getchar()) != '1' && o != '0'); return o;

    }

}

using namespace IO;



char Map[233];



int main() {

	Map['b'] = 'd';

	Map['d'] = 'b';

	Map['p'] = 'q';

	Map['q'] = 'p';



	string str; cin >> str;

	for (int pos = 0; pos <= str.size() / 2; pos++)

		if (Map[str[pos]] != str[str.size() - 1 - pos]) {

			puts("No"); return 0;

		}

	puts("Yes");

}


// Faster C++ solution for the same problem (measured speedup: 5.341228597525474x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<cstdio>

#include<iostream>

#include<cstring>

#include<string>

using namespace std;

char s[100001],s2[100010],s3[100001];

int flag=1,len,L,R;

int main(){

    cin>>s;

    len=strlen(s);

    for(int i=len-1;i>=0;i--){

        L=i,R=len-i-1;

        if(s[L]==s[R])

        flag=0;

        else {

            flag=1;

            break;

        }

    }

    if(!flag){

        cout<<"Yes"<<endl;

        return 0;

    }

    int cnt=0;

    for(int i=0;i<len;i++){

        s2[len-i-1]=s[i];

        if(s2[len-i-1]=='b')

        s2[len-i-1]='d';

        else if(s2[len-i-1]=='d')

        s2[len-i-1]='b';

        else if(s2[len-i-1]=='p')

        s2[len-i-1]='q';

        else if(s2[len-i-1]=='q')

        s2[len-i-1]='p';

        if(s2[len-i-1]==s[len-i-1])

         cnt=1;

         else

         {

         	cnt=0;

         	break;

		 }

    }

    if(cnt==1){

        cout<<"Yes"<<endl;

        return 0;

    }

    cout<<"No"<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.332542837954549x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<queue>

#include<vector>

#include<algorithm>

#include<map>

#include<stack>

#include<bits/stdc++.h>

using namespace std;

int read()

{

	int kk(0);

	bool fh=true;

	char ch=getchar();

	for(;ch<'0'||ch>'9';)

	{

		if(ch=='-')fh=false;

		ch=getchar();

	}

	for(;ch>='0'&&ch<='9';)

	{

		kk=(kk<<3)+(kk<<1)+(ch^48);

		ch=getchar();

	}

	return fh?kk:-kk;

}

long long readll()

{

	long long kk(0);

	bool fh=true;

	char ch=getchar();

	for(;ch<'0'||ch>'9';)

	{

		if(ch=='-')fh=false;

		ch=getchar();

	}

	for(;ch>='0'&&ch<='9';)

	{

		kk=(kk<<3)+(kk<<1)+(ch^48);

		ch=getchar();

	}

	return fh?kk:-kk;

}

#define INF 214748364

#define MOD 1000000007

#define MAXN

char s[100010],s1[100010];

map<char,char> mp;

int main()

{

	mp['d']='b';

	mp['b']='d';

	mp['q']='p';

	mp['p']='q';

	scanf("%s",s);

	int len=strlen(s);

	for(int i=0;i<len;i++)s1[len-i-1]=s[i];

	for(int i=0;i<len;i++)s1[i]=mp[s1[i]];

	if(strcmp(s,s1)==0)puts("Yes");

	else puts("No");

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.318418640618678x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

int main()

{

  string s;

  cin>>s;

  int n=s.size();

  int ok=0;

  for(int i=0; i<n; i++)

  {

    int j=n-i-1;

    if((s[i]=='q'&&s[j]=='p')||(s[i]=='b'&&s[j]=='d') ||(s[i]=='d'&&s[j]=='b')||(s[i]=='p'&&s[j]=='q')) {continue;}

    ok=1;

  }

  if(ok==0)

  {

    cout<<"Yes"<<endl;

    return 0;

  }

  ok=0;

  for(int i=0; i<n; i++)

  {

    if(s[i]=='p')s[i]=='q';

    if(s[i]=='q')s[i]=='p';

    if(s[i]=='b')s[i]=='d';

    if(s[i]=='d')s[i]=='b';

  }

  for(int i=0; i<n; i++)

  {

    int j=n-i-1;

    if((s[i]=='q'&&s[j]=='p')||(s[i]=='b'&&s[j]=='d') ||(s[i]=='d'&&s[j]=='b')||(s[i]=='p'&&s[j]=='q')) {continue;}

    ok=1;

  }

  if(ok==0)cout<<"Yes"<<endl;

  else cout<<"No";

  return 0;

}


// Faster C++ solution for the same problem (measured speedup: 5.359170760009426x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

	string a;

	cin>>a;

	string b=a;

	reverse(a.begin(),a.end());

	for(int i=0;i<a.size();++i)

	{

		if(a[i]=='b') a[i]='d';

		else if(a[i]=='d') a[i]='b';

		else if(a[i]=='p') a[i]='q';

		else if(a[i]=='q') a[i]='p';

	}

	if(a==b) cout<<"Yes";

	else cout<<"No";

}

// Faster C++ solution for the same problem (measured speedup: 5.360757793043199x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{

    string a,b;

    cin>>a;

    b=a;

    reverse(a.begin(),a.end());

    for(int i=0;i<a.size();++i)

    {

        if(a[i]=='b') a[i]='d';

        else if(a[i]=='d') a[i]='b';

        else if(a[i]=='p') a[i]='q';

        else if(a[i]=='q') a[i]='p';

    }

    if(a==b) printf("Yes");

    else printf("No");

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.354228018409972x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
// =================================

//   author: memset0

//   date: 2018.12.25 22:11:27

//   website: https://memset0.cn/

// =================================

#include <bits/stdc++.h>

#define rep(i, l, r) for (register int i = l; i <= r; i++)

namespace ringo {

typedef long long ll;

typedef unsigned long long ull;

template <class T> inline void read(T &x) {

	x = 0; register char c = getchar(); register bool f = 0;

	while (!isdigit(c)) f ^= c == '-', c = getchar();

	while (isdigit(c)) x = x * 10 + c - '0', c = getchar();

	if (f) x = -x;

}

template <class T> inline void print(T x) {

	if (x < 0) putchar('-'), x = -x;

	if (x > 9) print(x / 10);

	putchar('0' + x % 10);

}

template <class T> inline void maxd(T &a, T b) { if (b > a) a = b; }

template <class T> inline void mind(T &a, T b) { if (b < a) a = b; }

template <class T> inline void print(T x, char c) { print(x), putchar(c); }

template <class T> inline T abs(const T &a) { if (a < 0) return -a; return a; }



const int N = 1e5 + 10;

int n, m, a[N], b[N];

char s[N];



void main() {

	scanf("%s", s + 1), n = strlen(s + 1);

	for (int i = 1; i <= n; i++)

		switch (s[i]) {

			case 'b': a[i] = 0; break;

			case 'p': a[i] = 1; break;

			case 'd': a[i] = 2; break;

			case 'q': a[i] = 3; break;

		}

	for (int i = 1; i <= n; i++) b[i] = a[i];

	std::reverse(b + 1, b + n + 1);

	for (int i = 1; i <= n; i++) b[i] = (b[i] + 2) % 4;

	for (int i = 1; i <= n; i++)

		if (a[i] != b[i]) {

			puts("No");

			return;

		}

	puts("Yes");

}



} signed main() { return ringo::main(), 0; }

// Faster C++ solution for the same problem (measured speedup: 5.265969417929571x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include <bits/stdc++.h>

#define rep(i,a,b) for(int i=a;i<=b;i++)

using namespace std;

string s;

int main(){

    cin>>s;

    string s1(s);

    reverse(s.begin(),s.end());

    rep(i,0,(int)s.size()-1){

        if(s1[i]=='b')s1[i]='d';

        else if(s1[i]=='d')s1[i]='b';

        else if(s1[i]=='p')s1[i]='q';

        else if(s1[i]=='q')s1[i]='p';

    }

    if(s1==s)printf("Yes");

    else printf("No");

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.308160831461912x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

char s[100005],ss[100005];

signed main(){

	scanf("%s",s);

	int len=strlen(s);

	for(int i=0;i<len;i++){

		switch(s[i]){

			case 'b':{ss[len-i-1]='d';break;}

			case 'd':{ss[len-i-1]='b';break;}

			case 'p':{ss[len-i-1]='q';break;}

			case 'q':{ss[len-i-1]='p';break;}

		}

	}

	if(strcmp(s,ss))printf("No");

	else printf("Yes");

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.268608959595928x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

char a[100010],b[100010];

int main(void){

    scanf("%s",a+1);

    int cnt=0;

    for(int i=strlen(a+1);i>=1;i--){

        b[++cnt]=a[i];

    }

    for(int i=1;i<=strlen(a+1);i++){

        if(a[i]=='b') a[i]='d';

        else if(a[i]=='d') a[i]='b';

        else if(a[i]=='p') a[i]='q';

        else if(a[i]=='q') a[i]='p';

    }

    for(int i=1;i<=strlen(a+1);i++){

        if(a[i]==b[i]) continue;

        else{

            printf("No\n");

            return 0;

        }

    }

    printf("Yes\n");

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.269492878244457x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

string s,b;

int main(){

	cin>>s;

	b=s;reverse(s.begin(),s.end());

	for(int i=0;i<s.size();i++)

		switch(s[i]){

			case 'b':s[i]='d';break;

			case 'd':s[i]='b';break;

			case 'p':s[i]='q';break;

			case 'q':s[i]='p';break;

		}

	s==b?puts("Yes"):puts("No");

	return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.310044749317112x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main()

{

    string a;

    cin>>a;

    string b=a;

    reverse(a.begin(),a.end());

    for(int i=0;i<a.size();++i)

        if (a[i]=='b') a[i]='d'; else

            if (a[i]=='d') a[i]='b'; else

                if (a[i]=='p') a[i]='q'; else

                    if (a[i]=='q') a[i]='p';

    if (a==b) printf("Yes"); else printf("No");

}

// Faster C++ solution for the same problem (measured speedup: 5.30799571416509x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

    string a;

    cin>>a;

    string b=a;

    reverse(a.begin(),a.end());//交换

    for(int i=0;i<a.size();++i)

    {

        if(a[i]=='b') a[i]='d';//镜面对称

        else if(a[i]=='d') a[i]='b';//请注意,else一定要加,不然就白做了

        else if(a[i]=='p') a[i]='q';

        else if(a[i]=='q') a[i]='p';

    }

    if(a==b) cout<<"Yes";//判断

    else cout<<"No";

}

// Faster C++ solution for the same problem (measured speedup: 5.360757793043199x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}
// Slower C++ solution:
#include<bits/stdc++.h>

using namespace std;

int main(){

    string a;

    cin>>a;

    string b=a;

    reverse(a.begin(),a.end());

    for(int i=0;i<a.size();++i)

    {

        if(a[i]=='b') a[i]='d';

        else if(a[i]=='d') a[i]='b';

        else if(a[i]=='p') a[i]='q';

        else if(a[i]=='q') a[i]='p';

    }

    if(a==b) cout<<"Yes"<<endl;

    else cout<<"No"<<endl;

    return 0;

}

// Faster C++ solution for the same problem (measured speedup: 5.373008834860594x):
#include<algorithm>

#include<cstring>

#include<cstdio>

#include<cmath>

char s[100007];

int get[128];

int main()

{

    get['b'] = 0 , get['p'] = 1 , get['d'] = 2 , get['q'] = 3;

    scanf("%s",s+1); // 从 s[1] 开始的

    for(int l = strlen(s+1), i = 1; i <= (l>>1) + 1; ++i)

    {

        if(get[s[i]]^((get[s[l-i+1]]+2)%4)) return printf("No\n"),0;

    } // 这个 +2 %4 刚好让 b 转到 d 上， d 转到 b 上， p 转到 q 上， q 转到 p 上 

    return printf("Yes\n"),0;

}