//Solution to Periodic Points in Piecewise Linear Maps
//By Luis Hernandez Corbato
//Complexity: O(m^3·log(n) + m·n)
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <ctime>
using namespace std;

#define MAX 80
int module;

vector <vector <int> > matrixproduct (vector <vector <int> > a, vector <vector <int> > b){
  int m = a.size();  
  vector <vector <int> > ret (m, vector<int>(m));
  for (int i = 0; i < m; i++)
    for (int j = 0; j < m; j++)
      for (int k = 0; k < m; k++){
        ret[i][j] += a[i][k] * b[k][j];
        ret[i][j] %= module;
      }
  return ret;
}

vector <vector <int> > exp (vector <vector <int> > a, int n){
  //computes a^n in O(logn) multiplications
  if (n == 1) return a; 
  vector <vector <int> > ret, b;
  b = exp(a, n / 2);
  ret = matrixproduct(b, b);
  if (n % 2 != 0) ret = matrixproduct(ret, a);
  return ret;
}

vector <int> compose (vector <int> a, vector <int> b){
  int m = a.size();
  vector <int> ret(m);
  for (int i = 0; i < m; i++){
    if (b[i] == -1) ret[i] = -1;
    else ret[i] = a[b[i]];
  }
  return ret;
}

vector <int> iterate (vector <int> a, int n){
  if (n == 1) return a; 
  vector <int> ret, b;
  b = iterate(a, n / 2);
  ret = compose(b, b);
  if (n % 2 != 0) ret = compose(ret, a);
  return ret;
}


int main (){

clock_t start, finish;

start = clock();

for (;;){
   //input
   int m, f[MAX + 1];
   cin >> m;
   if (!m) break;
   for (int i = 0; i <= m; i++) cin >> f[i];  
   int n;
   cin >> n >> module;
   
   vector <vector <int> > a(m, vector<int>(m));
   
   for (int i = 0; i < m; i++)
     for (int j = 0; j < m; j++)
       a[i][j] = 0LL;
       
   for (int i = 0; i < m; i++){
     int x = f[i];
     int y = f[i+1];    
     if (x > y) {int z = y; y = x; x = z;}
     for (int j = x; j < y; j++)
       a[j][i] = 1LL;
   }
   
   a = exp(a, n);
   //the trace of this matrix will be the output after minor
   //changes concerning points of the form i/m and their neighborhoods 
   
   int ret = 0;
   for (int i = 0; i < m; i++)
     ret = (ret + a[i][i]) % module;

   vector <int> b(3 * (m + 1));

   for (int i = 0; i < 3 * (m + 1); i++){
     int d = (i % 3) - 1;
     int k = i / 3;
     int dir = 0;
     if (k + d >= 0 && k + d <= m){
       if (f[k + d] > f[k]) dir = 1;
       else if (f[k + d] < f[k]) dir = -1;
      }
     b[i] = 3 * f[k] + 1 + dir;
    }

   b = iterate(b, n);
   //b[3*i+1] = 3*j+1 iff f^n(i/m) = j/m

   for (int i = 0; i < 3 * (m + 1); i++)
     if (i > 0 && i <= 3 * m && (i % 3) != 1)
        if (b[i] == i) ret = (ret + module - 1) % module;

   for (int i = 0; i <= m; i++)
     if (b[3 * i + 1] == 3 * i + 1) ret = (ret + 1) % module;

   //decide whether there is an infinite number of solutions
   bool infinite = false;
   for (int i = 0; i < m && !infinite; i++){
     bool inf = true;
     int j, p = 0;
     int interval = i;
     int orientation = 1;
     for (j = 1; j <= n && inf && !p; j++){
       if (abs(f[interval] - f[interval+1]) != 1) inf = false;
       interval = min(f[interval], f[interval+1]);
       if (f[interval] > f[interval + 1]) orientation *= -1;
       if (interval == i) p = j; 
     }
     if (orientation < 0) p = 2 * p;
     if (inf && p && (n % p == 0)) infinite = true;
   }
  
   if (infinite){
     cout << "Infinity" << endl;
     continue;
   }  

   cout << ret << endl;
  
 }

finish = clock();

cerr << "Tiempo: " << (double)(finish-start)/CLOCKS_PER_SEC << endl;

 return 0;
}