A: IP Address (Internet Protocol Address)

problem

Since one number string is given, find the number of valid delimiters for IPv4 IP addresses. However, the effective delimiters for IPv4 IP addresses are as follows.

* The sequence of numbers is divided into four, and each of the separated columns satisfies all of the following.
* Not an empty string.
* When read as a decimal number, it is an integer between 0 and 255.
* If the number is 0, then the column is `0` (` 00` etc. are not allowed).
* If the number is non-zero, the first digit is not `0` (`01` etc. are not allowed).



Input format


S

Constraint

* 4 \ leq | S | \ leq 12
* S contains only numbers.



Output format

Output the number on one line according to the valid delimiter.

Input example 1


123456789

Output example 1


1

There is one valid delimiter, `123.45.67.89`.

Input example 2


768426

Output example 2


6





Example

Input

123456789


Output

1
#include <bits/stdc++.h>
#define GET_MACRO(_1,_2,_3,_4,_5,_6,_7,_8,NAME,...) NAME
#define pr(...) cerr<< GET_MACRO(__VA_ARGS__,pr8,pr7,pr6,pr5,pr4,pr3,pr2,pr1)(__VA_ARGS__) <<endl
#define pr1(a) (#a)<<"="<<(a)<<" "
#define pr2(a,b) pr1(a)<<pr1(b)
#define pr3(a,b,c) pr1(a)<<pr2(b,c)
#define pr4(a,b,c,d) pr1(a)<<pr3(b,c,d)
#define pr5(a,b,c,d,e) pr1(a)<<pr4(b,c,d,e)
#define pr6(a,b,c,d,e,f) pr1(a)<<pr5(b,c,d,e,f)
#define pr7(a,b,c,d,e,f,g) pr1(a)<<pr6(b,c,d,e,f,g)
#define pr8(a,b,c,d,e,f,g,h) pr1(a)<<pr7(b,c,d,e,f,g,h)
using namespace std;
using Int = long long;
using _int = int;
using ll = long long;
using Double = long double;
const Int INF = (1LL<<55)+1e9; // ~ 3.6 * 1e16
const Int mod = (1e9)+7;
const Double EPS = 1e-8;
const Double PI = 6.0 * asin((Double)0.5);
using P = pair<Int,Int>;
using T = tuple<Int,Int,Int>;
template<class T> T Max(T &a,T b){return a=max(a,b);}
template<class T> T Min(T &a,T b){return a=min(a,b);}
template<class T1, class T2> ostream& operator<<(ostream& o,pair<T1,T2> p){return o<<"("<<p.first<<","<<p.second<<")";}
template<class T1, class T2, class T3> ostream& operator<<(ostream& o,tuple<T1,T2,T3> t){
  return o<<"("<<get<0>(t)<<","<<get<1>(t)<<","<<get<2>(t)<<")";}
template<class T1, class T2> istream& operator>>(istream& i,pair<T1,T2> &p){return i>>p.first>>p.second;}
template<class T> ostream& operator<<(ostream& o,vector<T> &a){Int i=0;for(T t:a)o<<(i++?" ":"")<<t;return o;}
template<class T> istream& operator>>(istream& i,vector<T> &a){for(T &t:a)i>>t;return i;}
template<class T>void prArr(T a,string s=" "){Int i=0;for(auto t:a)cout<<(i++?s:"")<<t;cout<<endl;}

string s;
int dfs(int idx, int cnt){
  if(idx == (int)s.size()) return cnt == 4;
  if(cnt > 4) return 0;
  
  int res = 0;
  string val;
  int len = 0;
  for(int i=idx;i<(int)s.size();i++){
    if(len++ > 5) break; 
    val += s[i];
    if(val == "0") {res += dfs(i+1, cnt+1); break;}
    if(1 <= stoi(val) && stoi(val) <= 255) res += dfs(i+1, cnt + 1);
  }
  return res;
}

signed main(){
  srand((unsigned)time(NULL));
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  cout << fixed << setprecision(12);

  cin>>s;

  int ans = dfs(0, 0);
  cout<<ans<<endl;
  return 0;
}

