There live square people in a square country. Everything in this country is square also. Thus, the Square Parliament has passed a law about a land. According to the law each citizen of the country has a right to buy land. A land is sold in squares, surely. Moreover, a length of a square side must be a positive integer amount of meters. Buying a square of land with a side a one pays a 2 quadrics (a local currency) and gets a square certificate of a landowner.

One citizen of the country has decided to invest all of his N quadrics into the land. He can, surely, do it, buying square pieces 1 × 1 meters. At the same time the citizen has requested to minimize an amount of pieces he buys: "It will be easier for me to pay taxes," — he has said. He has bought the land successfully.

Your task is to find out a number of certificates he has gotten.

Input

The only line contains a positive integer N ≤ 60 000 , that is a number of quadrics that the citizen has invested.

Output

The only line contains a number of certificates that he has gotten.

Example

344

3

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

int main()
{
    int w[61234],dp[61234];
    int n,i,j;
    for(i=1;i<=245;i++)
    {
        w[i]=i*i;
    }
    while(cin >> n)
    {
        memset(dp,0,sizeof(dp));
        int v=sqrt(n);
        for(i=w[1];i<=n;i++)
        {
            dp[i]=i;
        }
        for(i=2;i<=v;i++)
        {
            for(j=w[i];j<=n;j++)
            {
                dp[j]=min(dp[j],dp[j-w[i]]+1);
            }
        }
        cout << dp[n] << endl;
    }
    return 0;
}

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

int dp[60000+100];

int dfs(int a)
{
    if(a==0)
    {
        return 0;
    }
    int &ans=dp[a];
    if(ans!=-1)
    {
        return ans;
    }
    ans=1e9;
    for(int i=1; i*i<=a; i++)
    {
        ans=min(ans,dfs(a-i*i)+1);
    }
    return ans;
}

int main()
{
    int n;
    memset(dp,-1,sizeof(dp));
    while(cin >> n)
    {
        cout << dfs(n) << endl;
    }
    return 0;
}