1 /**************************************************************
2 Problem: 3285
3 User: idy002
4 Language: C++
5 Result: Accepted
6 Time:756 ms
7 Memory:32072 kb
8 ****************************************************************/
9
10 #include <cstdio>
11 #include <cmath>
12 #include <cstring>
13 #include <cctype>
14 #define N 1000010
15
16 typedef long long dnt;
17
18 const int Hmod = 60793;
19 struct Hash {
20 int head[N], key[N], val[N], next[N], etot;
21 void init() {
22 etot = 0;
23 memset( head, 0, sizeof(head) );
24 }
25 void insert( int k, int v ) {
26 int kk = k%Hmod;
27 etot++;
28 key[etot] = k;
29 val[etot] = v;
30 next[etot] = head[kk];
31 head[kk] = etot;
32 }
33 int query( int k ) {
34 int kk = k%Hmod;
35 for( int t=head[kk]; t; t=next[t] )
36 if( key[t]==k ) return val[t];
37 return -1;
38 }
39 }hash;
40
41 dnt mpow( dnt a, int b, int c ) {
42 dnt rt;
43 for( rt=1; b; b>>=1,a=(a*a)%c )
44 if( b&1 ) rt=(rt*a)%c;
45 return rt;
46 }
47 int findroot( int p ) {
48 int phi = p-1;
49 int tmp = phi;
50 int stk[50], top;
51 top = 0;
52 for( int i=2; i<=(1<<16); i++ ) {
53 if( tmp%i==0 ) {
54 stk[++top] = i;
55 do {
56 tmp/=i;
57 }while( tmp%i==0 );
58 }
59 }
60 if( tmp!=1 )
61 stk[++top] = tmp;
62 for( int r=1; ; r++ ) {
63 bool ok = true;
64 for( int i=1; i<=top; i++ ) {
65 if( mpow(r,phi/stk[i],p)==1 ) {
66 ok=false;
67 break;
68 }
69 }
70 if( ok ) return r;
71 }
72 }
73 dnt ind( dnt r, int a, int p ) { // ind_r(a) mod p-1
74 int m = ceil(sqrt(p-1));
75 hash.init();
76 dnt cur = 1;
77 for( int i=0; i<m; i++ ) {
78 if( cur==a ) return i;
79 hash.insert( cur, i );
80 cur = (cur*r) % p;
81 }
82 dnt base;
83 base = cur = mpow(cur,p-2,p);
84 for( int i=m; i<p; i+=m,cur=(cur*base)%p ) {
85 int j = hash.query( a*cur%p );
86 if( j!=-1 ) return i+j;
87 }
88 return -1; // impossible
89 }
90 dnt gcd( dnt a, dnt b ) {
91 return b ? gcd(b,a%b) : a;
92 }
93 void exgcd( dnt a, dnt b, dnt &d, dnt &x, dnt &y ) {
94 if( b==0 ) {
95 d=a, x=1, y=0;
96 } else {
97 exgcd(b,a%b,d,y,x);
98 y-=a/b*x;
99 }
100 }
101 dnt meq( dnt a, dnt b, dnt c ) { // ax=b mod c
102 dnt d, dd, x, y;
103 a = (a%c+c)%c;
104 b = (b%c+c)%c;
105 d = gcd(a,c);
106 if( b%d!=0 ) return -1;
107 exgcd(a/d,c/d,dd,x,y);
108 x = x*(b/d);
109 x = (x%(c/d)+(c/d))%(c/d);
110 if( x==0 ) x+=c/d;
111 return x;
112 }
113
114 dnt a, b, c, g, p, r;
115 int aa[N], bb[N], cc[N], gg[N];
116
117 void read( int a[] ) {
118 int i;
119 char ch;
120 for( i=0; isdigit(ch=getchar()); i++ )
121 a[i] = ch-'0';
122 a[i] = -1;
123 }
124 dnt modulo( int a[], dnt mod ) {
125 dnt rt = 0;
126 for( int i=0; a[i]!=-1; i++ )
127 rt = (rt*10 + a[i]) % mod;
128 return rt;
129 }
130 int main() {
131 read(aa);
132 read(bb);
133 read(cc);
134 read(gg);
135 scanf( "%lld", &p );
136 a = modulo(aa,p-1);
137 b = modulo(bb,p-1),
138 c = modulo(cc,p);
139 g = modulo(gg,p);
140
141 if( g%p==0 || c%p==0 ) {
142 if( g%p==0 && c%p==0 )
143 printf( "1\n" );
144 else
145 printf( "no solution\n" );
146 return 0;
147 }
148 r = findroot(p);
149 // fprintf( stderr, "%d\n", (int)r );
150 dnt ans = meq( a*ind(r,g,p), ind(r,c,p)-b*ind(r,g,p), p-1 );
151 if( ans<0 )
152 printf( "no solution\n" );
153 else
154 printf( "%lld\n", ans );
155 }
