The HR manager was disappointed again. The last applicant failed the interview the same way as 24 previous ones. "Do I give such a hard task?" — the HR manager thought. "Just raise number 5 to the power of n and get last two digits of the number. Yes, of course, n can be rather big, and one cannot find the power using a calculator, but we need people who are able to think, not just follow the instructions."

Could you pass the interview in the machine vision company in IT City?

Input

The only line of the input contains a single integer n (2 ≤ n ≤ 2·1018) — the power in which you need to raise number 5.

Output

Output the last two digits of 5n without spaces between them.

Examples

Input

2


Output

25
7
 2000000000000000000
 1000000000000000000
 987654321012345678

25
 25
 25
 25

#include<bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp> // Common file 
#include <ext/pb_ds/tree_policy.hpp> 
#include <functional> // for less 
using namespace __gnu_pbds; 

////////////////////////////////////////////////////////

typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ldb;
 
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;

static mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
		

//////////////////////////////////////////////////////

#define pb push_back
#define pf push_front
#define popb pop_back()
#define popf pop_front()



#define rep(i, m, n) for (int i = m; i < n; i++)
#define repi(i, n) for (int i = n - 1; i >= 0; i--)


#define data_set tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>
#define set_data tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
/*		 data_set p;
		   p.insert(var);
		   1)value at i th index in sorted array = *p.find_by_order(i);
		   2)index of num in array = p.order_of_key(num);
		   3)if x is not in array ,pos if included = p.order_of_key(x);
*/

#define deci(n) fixed << setprecision(n) 
#define high(n) __builtin_popcount(n)			 //no. of ones in set bits
#define highll(n) __builtin_popcountll(n)		 //no. of ones in set bits (ll)
#define parity(n) __builtin_parity(n)			//odd parity->true else false
#define clz(n) __builtin_clz(n)					//clz = count leading zero’s
#define clzll(n) __builtin_clzll(n)				//clz = count leading zero’s (ll)
#define ctz(n) __builtin_ctz(n)					//ctz = count trailing zeros
#define all(v) v.B,v.E
#define S size()
#define E end()
#define B begin()
#define L length()

#define mod1 1000000007
#define mod2 998244353
#define INF LLONG_MAX
#define PI 3.1415926535897932384


#define gcd __gcd
#define search binary_search					//bool
#define lol ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define endl "\n"

////////////////////////////////////////////////////////////////////////

inline ll power(ll x, ll y, ll p = mod2)
{
    ll res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}
//////////////////////////////////////////////

inline ll modadd(ll x, ll y, ll m = mod2)
{
    x += y;
    if (x >= m)
        x -= m;
    return x;
}

/////////////////////////////////////////////

inline ll modmul(ll x, ll y, ll m = mod2)
{
    return ((x % m)*(y % m))%m;
}
/////////////////////////////////////////////
bool isprime(ll n)
{
    ll flagi = 1;
    for (ll i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) { 
            flagi = 0; 
            break; 
        } 
    } 
    if(n<=1){
        flagi=0;
    }else if(n==2){
        flagi=1; 
    }
    
    if(flagi==1){
        return true;
    }else{
        return false;
    }
    
}
////////////////////////////////////////////
ll arraySum(ll a[], ll n)  
{ 
    int initial_sum  = 0;  
    return accumulate(a, a+n, initial_sum); 
} 

int main(){
	lol;

	    ll n;
	    cin>>n;
	    cout<<pow(5,n)<<endl;


}