Problem: Many people are aware of DMCA – Digital Millennium Copyright Act. But another recently proposed DMCA – Digital Millennium Calculation Act – is much less known.

In this problem you need to find a root of a number according to this new DMCA law.

Input

The input contains a single integer a (1 ≤ a ≤ 1000000).

Output

Output the result – an integer number.

Examples

Input


1


Output


1


Input


81


Output


9

Solution (in CPP):

Solution:
#include <iostream>
#include<bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
//#define endl "\n"
#define vi vector<int>
#define vii vector<pair<int,int>>
#define pb push_back
using namespace std;
using ll = long long;
const ll mod = 1e9+7;

ll fac[1000001];

void pre(){
    fac[0]=fac[1]=1;
    for(int i=2;i<=1000001;i++)
{

    fac[i]=(fac[i-1]*1LL*i)%mod;
}
}

ll binpower(ll a,ll n){
    ll res=1;
    while(n){

    if(n%2)
        res = (res*1LL*a)%mod;

        n/=2;
        a=(a*1LL*a)%mod;

    }
    return res;

}

ll nCrmod(ll n,ll r){


ll res= fac[n];
res = (res*1LL*binpower(fac[r],mod-2))%mod;
res= (res*1LL*binpower(fac[n-r],mod-2))%mod;
return res;

}

long long ncr(int n, int r) {
    if(r > n - r) r = n - r; // because C(n, r) == C(n, n - r)
    long long ans = 1;
    int i;

    for(i = 1; i <= r; i++) {
        ans *= n - r + i;
        ans /= i;
    }

    return ans;
}
ll modexp(ll a,ll b,ll m){
    if(b==0)return 1;
    if(b%2==0){
        ll y =  modexp(a,b/2,m);
        return (y*y)%m;
    }
    else{
        return ((a%m)*modexp(a,b-1,m))%m;
    }
}

ll modinv(ll a,ll m){
return modexp(a,m-2,m);
//works only for prime m.
//else calculate phi(m)-1.
}

void SieveOfEratosthenes(ll n)
{
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be false if i is Not a prime, else true.
    bool prime[n+1];
    memset(prime, true, sizeof(prime));

    for (ll p=2; p*p<=n; p++)
    {
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == true)
        {
            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked.
            for (ll i=p*p; i<=n; i += p)
                prime[i] = false;
        }
    }

    // Print all prime numbers
    for (ll p=2; p<=n; p++)
       if (prime[p])
          cout << p << " ";
}

vector<int> Centroid(vector<int>g[],int n) {

        vector<int> centroid;
        vector<int> sz(n+1);
        function<void (int, int)> dfs = [&](int u, int prev) {
                sz[u] = 1;
                bool is_centroid = true;
                for (auto v : g[u]) if (v != prev) {
                        dfs(v, u);
                        sz[u] += sz[v];
                        if (sz[v] > n / 2) is_centroid = false;
                }
                if (n - sz[u] > n / 2) is_centroid = false;
                if (is_centroid) centroid.push_back(u);
        };
        dfs(1, 0);
        return centroid;
}


int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll p;
    cin>>p;

    while(1){
            if(p/10==0)break;
    ll q=0;
    while(p>0){
        q+=p%10;
        p/=10;
    }
    p=q;
    }
cout<<p<<endl;

    return 0;
}