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<deque>
#include<queue>
#include<vector>
#include<algorithm>
#include<iostream>
#include<set>
#include<cmath>
#include<tuple>
#include<string>
#include<chrono>
#include<functional>
#include<iterator>
#include<random>
#include<unordered_set>
#include<array>
#include<map>
#include<iomanip>
#include<assert.h>
#include<bitset>
#include<stack>
#include<memory>
using namespace std;
typedef long long int llint;
typedef long double lldo;
#define mp make_pair
#define mt make_tuple
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define fir first
#define sec second
#define res resize
#define ins insert
#define era erase
/*
cout<<fixed<<setprecision(20);
cin.tie(0);
ios::sync_with_stdio(false);
*/
const llint mod=1000000007;
const llint big=2.19e18+1;
const long double pai=3.141592653589793238462643383279502884197;
const long double eps=1e-15;
template <class T,class U>bool mineq(T& a,U b){if(a>b){a=b;return true;}return false;}
template <class T,class U>bool maxeq(T& a,U b){if(a<b){a=b;return true;}return false;}
llint gcd(llint a,llint b){if(a%b==0){return b;}else return gcd(b,a%b);}
llint lcm(llint a,llint b){if(a==0){return b;}return a/gcd(a,b)*b;}
template<class T> void SO(T& ve){sort(ve.begin(),ve.end());}
template<class T> void REV(T& ve){reverse(ve.begin(),ve.end());}
template<class T>llint LBI(vector<T>&ar,T in){return lower_bound(ar.begin(),ar.end(),in)-ar.begin();}
template<class T>llint UBI(vector<T>&ar,T in){return upper_bound(ar.begin(),ar.end(),in)-ar.begin();}
int main(void){
	string str;cin>>str;
	int dp[5][13]={0};
	dp[0][0]=1;
	int i,j,k;
	for(i=1;i<=4;i++){
		for(j=1;j<=str.size();j++){
			int suu=0,kak=1;
			for(k=j-1;k>=0;k--){
				suu+=(str[k]-'0')*kak;
				if(suu>255){break;}
				kak*=10;
				if(str[k]=='0'&&kak>10){continue;}
				dp[i][j]+=dp[i-1][k];
			}
		}
	}
	cout<<dp[4][str.size()]<<endl;
}
