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#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <cmath>
#include <cstring>
#include <string>
#include <iostream>
#include <complex>
#include <sstream>
#include <cassert>
using namespace std;
 
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
typedef vector<int> VI;
typedef pair<int,int> PII;
 
#define REP(i,n) for(int i=0;i<(n);++i)
#define SIZE(c) ((int)((c).size()))
#define FOR(i,a,b) for (int i=(a); i<(b); ++i)
#define FOREACH(i,x) for (__typeof((x).begin()) i=(x).begin(); i!=(x).end(); ++i)
#define FORD(i,a,b) for (int i=(a)-1; i>=(b); --i)
#define ALL(v) (v).begin(), (v).end()
 
#define pb push_back
#define mp make_pair
#define st first
#define nd second

LL p10[30];
LL pisano[30];

char C[30];
vector<LL> candidates[30];

LL mul(LL a, LL b) {
  LL a0 = a % p10[9], a1 = a / p10[9];
  LL b0 = b % p10[9], b1 = b / p10[9];
  
  LL res = a0 * b0;
  res += ((a1 * b0 + a0 * b1) % p10[9]) * p10[9];  
  return res % p10[18];
}

map<LL, LL> F;
LL fib(LL N) {
  if (N < 3) return min(N, 1LL);
  if (F.find(N) != F.end()) return F[N];
  
  LL K = N / 2;
  if (N == 2 * K) {
    return F[2*K] = (mul(fib(K+1), fib(K+1)) - mul(fib(K-1), fib(K-1)) + p10[18]) % p10[18];
  } else {
    return F[2*K+1] = (mul(fib(K+1), fib(K+1)) + mul(fib(K), fib(K)) + p10[18]) % p10[18];
  }
}

int main() {
  p10[0] = 1;
  FOR(i,1,20) p10[i] = p10[i-1] * 10;

  pisano[0] = 1;
  pisano[1] = 60;
  pisano[2] = 300;
  pisano[3] = 1500;
  FOR(i,4,20) pisano[i] = pisano[i-1] * 10;

  scanf("%s", C);
  int N = strlen(C);
  reverse(C, C + N);
  REP(i,N) C[i] -= '0';

  candidates[0].pb(0);
  LL remainder = 0;
  FOR(i,1,N+1) {
    remainder += p10[i-1] * C[i-1];
    FOREACH(it, candidates[i-1]) {
      for (LL c = *it; c < pisano[i]; c += pisano[i-1]) {
        if (fib(c) % p10[i] == remainder) candidates[i].pb(c);
      }
    }
  }
  
  if (candidates[N].empty()) {
    printf("NIE\n");
  } else {
    printf("%lld\n", pisano[18] + candidates[N][0]);
  }
}