stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define rr register

#define int long long

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

using namespace std;

const int maxx=1e4;

int n,m;

string a,b;

signed main(){

    cin>>a;

    b=a;

    bool t=true;

    int len=a.length()-1;

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

        switch(a[i]){

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

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

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

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

        }

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

        if(a[i]!=b[len-i])

            t=false;

    if(t)

        cout<<"Yes"<<endl;

    else

        cout<<"No"<<endl;

        return 0;

}

// Optimized solution to the same problem (faster):
#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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#pragma GCC optimize(2)

#include<bits/stdc++.h>

using namespace std;

string s, t;

int main(){

    ios::sync_with_stdio(false);

    cin >> s;

    t = s;

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

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

        if(s[i] == 98) s[i] = 100;

        else if(s[i] == 100) s[i] = 98;

        else if(s[i] == 112) s[i] = 113;

        else if(s[i] == 113) s[i] = 112;

    }

    if(s == t) cout << (char)89 << (char)101 << (char)115;

    else cout << (char)78 << (char)111;

    return 0;

}

// Optimized solution to the same problem (faster):
#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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>



using namespace std;



char a[100010],b[100010];

int cc,flag=1;



void jud()

{

	for(int i=0;i<cc-1;i++)

	{

		if(a[i]!=b[cc-i-2])

		{

			flag=0;

			return ;

		}

	}

}



int main( )

{

	while(~scanf("%c",&a[cc]))

	{

		cc++;

	}

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

	{

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

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

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

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

	}

	jud();

	if(flag==1)cout<<"Yes"<<endl;

	else cout<<"No"<<endl;

	return 0;

}

// Optimized solution to the same problem (faster):
#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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#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";

}

// Optimized solution to the same problem (faster):
#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;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<cstdio>

#include<iostream>

#include<cstring>

#include<stack>

using namespace std;

stack<char>s;

string a;

int main(){

    cin>>a;

    int n=a.size();

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

        s.push(a[i]);

    }

    bool ju=1;

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

        char top=s.top();

        if(top=='b') top='d';

        else if(top=='d') top='b';

        else if(top=='p') top='q';

        else if(top=='q') top='p';

        if(top==a[i]) {

            ju=1;

            s.pop();

        }

        else {

            ju=0;

            break;

        }

    }

    if(ju==0) printf("No");

    else printf("Yes");

    return 0;

}

// Optimized solution to the same problem (faster):
#include <algorithm>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <string>

#include <queue>

#include <cmath>

#include <set>

#include <map>

const int oo=2139063143;

const int N=1010000;

const int P=1000000007;

using namespace std;

typedef long long LL;

typedef double db;

//char buf[1<<24],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;

//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

typedef int DATA1;

typedef int DATA2;

inline void sc (DATA1 &x)

{

    x=0; static int p; p=1; static char c; c=getchar();

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

    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }

    x*=p;

}

inline void print (DATA2 x)

{

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

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

	putchar(x%10+'0');

}

inline void pr (DATA2 x) { print(x),putchar('\n'); }

char change (char c)

{

	if (c=='b') return 'd';

	if (c=='d') return 'b';

	if (c=='p') return 'q';

	return 'p';

}

char s[N];

int main()

{

	//freopen (".in","r",stdin);

	//freopen (".out","w",stdout);

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

	int n=strlen (s+1);

	if (n&1) { puts("No"); return 0; }

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

		if (s[i]!=change (s[n-i+1])) { puts("No"); return 0; }

	puts("Yes");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <algorithm>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <string>

#include <queue>

#include <cmath>

#include <set>

#include <map>

const int oo=2139063143;

const int N=1010000;

const int P=1000000007;

using namespace std;

typedef long long LL;

typedef double db;

//char buf[1<<24],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;

//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

typedef int DATA1;

typedef int DATA2;

inline void sc (DATA1 &x)

{

    x=0; static int p; p=1; static char c; c=getchar();

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

    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }

    x*=p;

}

inline void print (DATA2 x)

{

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

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

	putchar(x%10+'0');

}

inline void pr (DATA2 x) { print(x),putchar('\n'); }

char change (char c)

{

	if (c=='b') return 'd';

	if (c=='d') return 'b';

	if (c=='p') return 'q';

	return 'p';

}

char s[N];

int main()

{

	//freopen (".in","r",stdin);

	//freopen (".out","w",stdout);

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

	int n=strlen (s+1);

	if (n&1) { puts("No"); return 0; }

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

		if (s[i]!=change (s[n-i+1])) { puts("No"); return 0; }

	puts("Yes");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <algorithm>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <string>

#include <queue>

#include <cmath>

#include <set>

#include <map>

const int oo=2139063143;

const int N=1010000;

const int P=1000000007;

using namespace std;

typedef long long LL;

typedef double db;

//char buf[1<<24],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;

//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

typedef int DATA1;

typedef int DATA2;

inline void sc (DATA1 &x)

{

    x=0; static int p; p=1; static char c; c=getchar();

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

    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }

    x*=p;

}

inline void print (DATA2 x)

{

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

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

	putchar(x%10+'0');

}

inline void pr (DATA2 x) { print(x),putchar('\n'); }

char change (char c)

{

	if (c=='b') return 'd';

	if (c=='d') return 'b';

	if (c=='p') return 'q';

	return 'p';

}

char s[N];

int main()

{

	//freopen (".in","r",stdin);

	//freopen (".out","w",stdout);

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

	int n=strlen (s+1);

	if (n&1) { puts("No"); return 0; }

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

		if (s[i]!=change (s[n-i+1])) { puts("No"); return 0; }

	puts("Yes");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <algorithm>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <string>

#include <queue>

#include <cmath>

#include <set>

#include <map>

const int oo=2139063143;

const int N=1010000;

const int P=1000000007;

using namespace std;

typedef long long LL;

typedef double db;

//char buf[1<<24],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;

//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

typedef int DATA1;

typedef int DATA2;

inline void sc (DATA1 &x)

{

    x=0; static int p; p=1; static char c; c=getchar();

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

    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }

    x*=p;

}

inline void print (DATA2 x)

{

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

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

	putchar(x%10+'0');

}

inline void pr (DATA2 x) { print(x),putchar('\n'); }

char change (char c)

{

	if (c=='b') return 'd';

	if (c=='d') return 'b';

	if (c=='p') return 'q';

	return 'p';

}

char s[N];

int main()

{

	//freopen (".in","r",stdin);

	//freopen (".out","w",stdout);

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

	int n=strlen (s+1);

	if (n&1) { puts("No"); return 0; }

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

		if (s[i]!=change (s[n-i+1])) { puts("No"); return 0; }

	puts("Yes");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <algorithm>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <string>

#include <queue>

#include <cmath>

#include <set>

#include <map>

const int oo=2139063143;

const int N=1010000;

const int P=1000000007;

using namespace std;

typedef long long LL;

typedef double db;

//char buf[1<<24],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;

//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

typedef int DATA1;

typedef int DATA2;

inline void sc (DATA1 &x)

{

    x=0; static int p; p=1; static char c; c=getchar();

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

    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }

    x*=p;

}

inline void print (DATA2 x)

{

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

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

	putchar(x%10+'0');

}

inline void pr (DATA2 x) { print(x),putchar('\n'); }

char change (char c)

{

	if (c=='b') return 'd';

	if (c=='d') return 'b';

	if (c=='p') return 'q';

	return 'p';

}

char s[N];

int main()

{

	//freopen (".in","r",stdin);

	//freopen (".out","w",stdout);

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

	int n=strlen (s+1);

	if (n&1) { puts("No"); return 0; }

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

		if (s[i]!=change (s[n-i+1])) { puts("No"); return 0; }

	puts("Yes");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <algorithm>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <string>

#include <queue>

#include <cmath>

#include <set>

#include <map>

const int oo=2139063143;

const int N=1010000;

const int P=1000000007;

using namespace std;

typedef long long LL;

typedef double db;

//char buf[1<<24],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;

//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

typedef int DATA1;

typedef int DATA2;

inline void sc (DATA1 &x)

{

    x=0; static int p; p=1; static char c; c=getchar();

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

    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }

    x*=p;

}

inline void print (DATA2 x)

{

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

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

	putchar(x%10+'0');

}

inline void pr (DATA2 x) { print(x),putchar('\n'); }

char change (char c)

{

	if (c=='b') return 'd';

	if (c=='d') return 'b';

	if (c=='p') return 'q';

	return 'p';

}

char s[N];

int main()

{

	//freopen (".in","r",stdin);

	//freopen (".out","w",stdout);

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

	int n=strlen (s+1);

	if (n&1) { puts("No"); return 0; }

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

		if (s[i]!=change (s[n-i+1])) { puts("No"); return 0; }

	puts("Yes");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#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");

}


// Optimized solution to the same problem (faster):
#include <algorithm>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <string>

#include <queue>

#include <cmath>

#include <set>

#include <map>

const int oo=2139063143;

const int N=1010000;

const int P=1000000007;

using namespace std;

typedef long long LL;

typedef double db;

//char buf[1<<24],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;

//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

typedef int DATA1;

typedef int DATA2;

inline void sc (DATA1 &x)

{

    x=0; static int p; p=1; static char c; c=getchar();

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

    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }

    x*=p;

}

inline void print (DATA2 x)

{

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

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

	putchar(x%10+'0');

}

inline void pr (DATA2 x) { print(x),putchar('\n'); }

char change (char c)

{

	if (c=='b') return 'd';

	if (c=='d') return 'b';

	if (c=='p') return 'q';

	return 'p';

}

char s[N];

int main()

{

	//freopen (".in","r",stdin);

	//freopen (".out","w",stdout);

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

	int n=strlen (s+1);

	if (n&1) { puts("No"); return 0; }

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

		if (s[i]!=change (s[n-i+1])) { puts("No"); return 0; }

	puts("Yes");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <algorithm>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <string>

#include <queue>

#include <cmath>

#include <set>

#include <map>

const int oo=2139063143;

const int N=1010000;

const int P=1000000007;

using namespace std;

typedef long long LL;

typedef double db;

//char buf[1<<24],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;

//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

typedef int DATA1;

typedef int DATA2;

inline void sc (DATA1 &x)

{

    x=0; static int p; p=1; static char c; c=getchar();

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

    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }

    x*=p;

}

inline void print (DATA2 x)

{

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

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

	putchar(x%10+'0');

}

inline void pr (DATA2 x) { print(x),putchar('\n'); }

char change (char c)

{

	if (c=='b') return 'd';

	if (c=='d') return 'b';

	if (c=='p') return 'q';

	return 'p';

}

char s[N];

int main()

{

	//freopen (".in","r",stdin);

	//freopen (".out","w",stdout);

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

	int n=strlen (s+1);

	if (n&1) { puts("No"); return 0; }

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

		if (s[i]!=change (s[n-i+1])) { puts("No"); return 0; }

	puts("Yes");

	return 0;

}


// Fastest known solution for this problem:
// Original C++ solution (slower):
#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;

}

// Optimized solution to the same problem (faster):
#include <algorithm>

#include <cstdlib>

#include <cstring>

#include <cstdio>

#include <string>

#include <queue>

#include <cmath>

#include <set>

#include <map>

const int oo=2139063143;

const int N=1010000;

const int P=1000000007;

using namespace std;

typedef long long LL;

typedef double db;

//char buf[1<<24],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;

//#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)

typedef int DATA1;

typedef int DATA2;

inline void sc (DATA1 &x)

{

    x=0; static int p; p=1; static char c; c=getchar();

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

    while ( isdigit(c)) { x=(x<<1)+(x<<3)+(c-48); c=getchar(); }

    x*=p;

}

inline void print (DATA2 x)

{

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

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

	putchar(x%10+'0');

}

inline void pr (DATA2 x) { print(x),putchar('\n'); }

char change (char c)

{

	if (c=='b') return 'd';

	if (c=='d') return 'b';

	if (c=='p') return 'q';

	return 'p';

}

char s[N];

int main()

{

	//freopen (".in","r",stdin);

	//freopen (".out","w",stdout);

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

	int n=strlen (s+1);

	if (n&1) { puts("No"); return 0; }

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

		if (s[i]!=change (s[n-i+1])) { puts("No"); return 0; }

	puts("Yes");

	return 0;

}


// Fastest known solution for this problem: