#include <bits/stdc++.h>
using namespace std;
int expresie(int &coefx, int &coefy, int &val);
int termen(int &coefx, int &coefy, int &val);
int factor(int &coefx, int &coefy, int &val);
char s[502],s0[502]; int p=-1;
bool ok[5000001];
int expresie(int &coefx, int &coefy, int &val){ int auxcoefx=0, auxcoefy=0, auxval=0; char semn;
p++;
termen(auxcoefx, auxcoefy, auxval);
coefx=auxcoefx; coefy=auxcoefy; val=auxval;
while(s[p]=='+' || s[p]=='-'){semn=s[p]; p++; termen(auxcoefx,auxcoefy,auxval);
 if(semn=='+'){ coefx+=auxcoefx; coefy+=auxcoefy; val+=auxval;}
 else {coefx-=auxcoefx; coefy-=auxcoefy; val-=auxval;}
}
}

int termen(int &coefx, int &coefy, int &val){ int auxcoefx=0, auxcoefy=0, auxval=0;
//aici inmultim;
if(s[p]=='x'){auxcoefx=1; auxcoefy=0; auxval=0; p++;}
  else if(s[p]=='y'){auxcoefx=0; auxcoefy=1; auxval=0; p++;}
  else factor(auxcoefx,auxcoefy,auxval);
  coefx=auxcoefx; coefy=auxcoefy; val=auxval;
while(s[p]=='*'){
    p++; if(s[p]=='x'){auxcoefx=1; auxcoefy=0; auxval=0; p++;}
else if(s[p]=='y'){auxcoefx=0; auxcoefy=1; auxval=0; p++;}
else factor(auxcoefx,auxcoefy,auxval);
//separare pe cazuri: avem inmultiri de forma (Ax+B)*valoare, valoare*(Ax+B) si analoage ptr y; sau pur si simplu doua numere naturale inmultite
if(coefx!=0)coefx*=auxval;
else if(auxcoefx!=0)coefx=auxcoefx*val;
if(coefy!=0)coefy*=auxval;
else if(auxcoefy!=0)coefy=auxcoefy*val;
val*=auxval;
}
}

int factor(int &coefx, int &coefy, int &val){ int auxcoefx=0, auxcoefy=0, auxval=0;
if(s[p]=='('){expresie(auxcoefx, auxcoefy, auxval); coefx=auxcoefx; coefy=auxcoefy; val=auxval; p++;}
else{
        int nr=0;
        while(isdigit(s[p])){nr=nr*10+(s[p]-'0'); p++;}
        val=nr;
}
}

int main()
{
    int coefx,coefy,val,meh1,meh2,val2,i=0,j,len0,len,poz,a,b,c,deverificat,nrsol=0,oky=0,okx,x,y; meh1=meh2=val2=coefx=coefy=val=0;
    cin.getline(s0,257);
    //separare Membru stang
    while(s0[i]!='='){s[i]=s0[i]; i++;}
//    cout<<s<<" ";
    poz=++i; len=strlen(s);
    //daca apare o structura de forma 12x, sau (-5)y, introducem "*" pentru a ne usura munca mai apoi
       //in plus, marchez daca apare y in expresie
    for(i=1;i<len;i++)
        if(strchr("xy",s[i])){
            if(isdigit(s[i-1]) || s[i-1]==')'){
              for(j=len-1;j>=i;j--)
                 s[j+1]=s[j];
              s[i]='*';
              len++;
            }
            if(s[i]=='y')oky=1;
        }
    expresie(coefx,coefy,val);
    //resetare s;
    for(i=0;i<len;i++)s[i]=0; p=-1; i=0;
    //reluare proces membru drept, de data fiind mai simplu
    while(s0[poz]){s[i]=s0[poz]; poz++; i++;} len=strlen(s);
    expresie(meh1,meh2,val2);
    //momentam avem coefx*x+coefy*y+val=val2, trecem in ec. de forma  coefx*x+coefy*y+val=val, sau a*x+b*y=c;
    c=val2-val; b=coefy; a=coefx;
//    cout<<coefx<<" "<<coefy<<" "<<val<<" "<<val2;
//    cout<<'\n'<<a<<" "<<b<<" "<<c<<"   "<<oky<<'\n';
//    migalosenii: verificam daca coeficientii lui x sau/si lui y sunt 0;
    if(!oky || b==0){
        if(a!=0){
        if((c%a)==0){ deverificat=c/a;
            if(deverificat<2)cout<<"No solution";
            for(i=2,okx=1;i*i<=deverificat;i++)if(deverificat%i==0){cout<<"No solution"; okx=0; break;}
            if(okx){cout<<"x="<<deverificat; if(oky)cout<<",y=2";}
        }
        else cout<<"No solution";
        }
        //daca coeficientul lui a este 0
        else {if(c==0 && !oky)cout<<"x=2";
        else if(c==0 && oky)cout<<"x=2,y=2";
        else cout<<"No solution";
        }
    }
    else if(oky && b!=0){
         //ciur! + nu mai verificam numerele divizibile cu 2, pentru a reduce la 5*10^6/2 numarul de posibilitati
    for(i=3;i*i<=5000000;i+=2)
        if(!ok[i])
            for(j=i*i;j<=5000000;j+=i)ok[j]=1;
    //luam fiecare x prim si verificam daca y=(c-a*x)/b e si el prim; sa nu uitam sa verificam daca y e par(!)
    //x=2
    if((c-2*a)%b==0 && (c-a*x)/b >=2){y=(c-2*a)/b; if(y==2 || (ok[y]==0 && y%2==1)){cout<<"x="<<2<<",y="<<y; nrsol=1;}   }
    //x prim, >2
    for(x=3;x<=5000000 && !nrsol;x+=2)
        if(!ok[x])
          if((c-a*x)%b == 0 && (c-a*x)/b >= 2) {
           y=(c-a*x)/b; if(y==2 || (ok[y]==0 && y%2==1)){cout<<"x="<<x<<",y="<<y; nrsol=1;}
          }
       if(!nrsol)cout<<"No solution";
    }
    return 0;
}