You are given a positive integer n, written without leading zeroes (for example, the number 04 is incorrect). 

In one operation you can delete any digit of the given integer so that the result remains a positive integer without leading zeros.

Determine the minimum number of operations that you need to consistently apply to the given integer n to make from it the square of some positive integer or report that it is impossible.

An integer x is the square of some positive integer if and only if x=y^2 for some positive integer y.

Input

The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{9}). The number is given without leading zeroes.

Output

If it is impossible to make the square of some positive integer from n, print -1. In the other case, print the minimal number of operations required to do it.

Examples

Input

8314


Output

2


Input

625


Output

0


Input

333


Output

-1

Note

In the first example we should delete from 8314 the digits 3 and 4. After that 8314 become equals to 81, which is the square of the integer 9.

In the second example the given 625 is the square of the integer 25, so you should not delete anything. 

In the third example it is impossible to make the square from 333, so the answer is -1.
        ////////////////////////////////////////////////////
        //                                                //
        //  For her who keeps the fire kindling in me...  //
        //                                                //
        ////////////////////////////////////////////////////


import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileOutputStream;
import java.io.IOException; 
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
// import java.util.Scanner;
// import java.util.Arrays;
// import java.util.ArrayList;
// import java.util.Collections; 
// import java.util.StringTokenizer;
import java.util.*;
import java.lang.Math;
import java.awt.Point;                            

import java.awt.geom.*;

public class Solution{

    static class Pair{
        int x;
        int y;
    }




    static class DisjointUnionSets { 
        int[] rank, parent, ele_count; 
        int n; 
        
        // Constructor 
        public DisjointUnionSets(int n) 
        { 
            rank = new int[n]; 
            parent = new int[n];
            ele_count = new int[n];
            this.n = n; 
            makeSet(); 
        } 
        
        // Creates n sets with single item in each 
        void makeSet() 
        { 
            for (int i = 0; i < n; i++) { 
                // Initially, all elements are in 
                // their own set. 
                parent[i] = i; 
                ele_count[i] = 1;
            } 
        } 
        
        // Returns representative of x's set 
        int find(int x) 
        { 
            // Finds the representative of the set 
            // that x is an element of 
            if (parent[x] != x) { 
                // if x is not the parent of itself 
                // Then x is not the representative of 
                // his set,

                parent[x] = find(parent[x]); 
                
                // so we recursively call Find on its parent 
                // and move i's node directly under the 
                // representative of this set 
            } 
        
            return parent[x]; 
        }



        int setSize(int x){
            int par = find(x);
            return ele_count[par];
        }
        


        // Unites the set that includes x and the set 
        // that includes x 
        void union(int x, int y) 
        { 
            // Find representatives of two sets 
            int xRoot = find(x), yRoot = find(y); 
        
            // Elements are in the same set, no need 
            // to unite anything. 
            if (xRoot == yRoot) 
                return; 
        
            // If x's rank is less than y's rank 
            if (rank[xRoot] < rank[yRoot]){ 
        
                // Then move x under y  so that depth 
                // of tree remains less 
                parent[xRoot] = yRoot;
            
                ele_count[yRoot] = ele_count[yRoot] + ele_count[xRoot];
        
            // Else if y's rank is less than x's rank 
            }else if (rank[yRoot] < rank[xRoot]) {
        
                // Then move y under x so that depth of 
                // tree remains less 
                parent[yRoot] = xRoot; 
                ele_count[xRoot] = ele_count[yRoot] + ele_count[xRoot];
            
            } else // if ranks are the same 
            { 
                // Then move y under x (doesn't matter 
                // which one goes where) 
                parent[yRoot] = xRoot; 
        
                ele_count[xRoot] = ele_count[yRoot] + ele_count[xRoot];

                // And increment the result tree's 
                // rank by 1 
                rank[xRoot] = rank[xRoot] + 1; 
            } 
        } 
    } 


    static long power(long x, long y, long p) 
    { 
        // Initialize result 
        long res = 1l;      
        
        // Update x if it is more   
        // than or equal to p 
        x = x % p;  
    
        while (y > 0) 
        { 
            // If y is odd, multiply x 
            // with result 
            if((y & 1)==1) 
                res = (res * x) % p; 
    
            // y must be even now 
            // y = y / 2 
            y = y >> 1;  
            x = (x * x) % p;  
        } 
        return res; 
    } 



static int countGreater(int arr[], int n, int k) 
{ 
int l = 0; 
int r = n - 1; 

// Stores the index of the left most element 
// from the array which is greater than k 
int leftGreater = n; 

// Finds number of elements greater than k 
while (l <= r) { 
    int m = l + (r - l) / 2; 

    // If mid element is greater than 
    // k update leftGreater and r 
    if (arr[m] > k) { 
        leftGreater = m; 
        r = m - 1; 
    } 

    // If mid element is less than 
    // or equal to k update l 
    else
        l = m + 1; 
} 

// Return the count of elements greater than k 
return (n - leftGreater); 
} 



static int countLessEqual(int arr[], int n, int k) 
{ 
int l = 0; 
int r = n - 1; 

// Stores the index of the left most element 
// from the array which is greater than k 
int leftGreater = n; 

// Finds number of elements greater than k 
while (l <= r) { 
    int m = l + (r - l) / 2; 

    // If mid element is greater than 
    // k update leftGreater and r 
    if (arr[m] > k) { 
        leftGreater = m; 
        r = m - 1; 
    } 

    // If mid element is less than 
    // or equal to k update l 
    else
        l = m + 1; 
} 

// Return the count of elements greater than k 
return (leftGreater); 
} 



    public static void main(String args[]){
        try{
            FastReader s = new FastReader();
            BufferedWriter w = new BufferedWriter(new OutputStreamWriter(System.out));
            int i, q, j, k, x, y, t, n, a, ptr, ind, count, diff, key, l, r, b, num, size, max, sec_max, digits, beg, end, begl, endr, ld, up, down, maxele, minele, case_count;
            long ans, sqrt, sum, maxup, maxdown, m, X, Y, Q, MOD;//, sum2;
            boolean flag, flag2;
            boolean l2r;


            double avg;
            char c ;
            String str;

            // t = s.nextInt();

            // for(; t>0; t--){
            
                TreeSet<Integer> set = new TreeSet<Integer>();

                for(i=1; (i*i) <= 2000000000; i++)
                    set.add(i*i);

                // n = s.nextInt();
                // digits = (int)Math.ceil(Math.log(n)/Math.log(10d));
                str = s.nextLine();
                char arr[];
                digits = str.length();
                arr = str.toCharArray();

                max = -1; 
                for (i = ((1<<digits)-1); i > 0; i--) 
                { 
                    // System.out.print("{ "); 


                    num = 0;    count = 0;      flag = false;
                    // Print current subset 
                    for (j = 0; j < digits; j++){ 

                        // (1<<j) is a number with jth bit 1 
                        // so when we 'and' them with the 
                        // subset number we get which numbers 
                        // are present in the subset and which 
                        // are not 
                        
                        if ((i & (1 << j)) > 0){
                            num = num*10 + (((int)(arr[j])) - ((int)'0'));
                            if(num == 0){
                                flag = true;
                                break;
                            }
                            count++;
                            // System.out.print(set[j] + " ");

                        }
                    }

                    if(flag == false &&  set.contains(num))
                        max = (int)Math.max(max, count);

                    // System.out.println("}"); 
                }

                if(max != -1)
                    max = digits-max;

                w.write((max) + "\n");
                    
            // }
            w.close();
        }catch(Exception e){
            e.printStackTrace();
        }
    }
}
                
                
//========================================### FAST IO ###=========================================//
class FastReader 
{ 
    BufferedReader br; 
    StringTokenizer st; 

    public FastReader() 
    { 
        br = new BufferedReader(new
                    InputStreamReader(System.in)); 
    } 

    String next() 
    { 
        while (st == null || !st.hasMoreElements()) 
        { 
            try
            { 
                st = new StringTokenizer(br.readLine()); 
            } 
            catch (IOException  e) 
            { 
                e.printStackTrace(); 
            } 
        } 
        return st.nextToken(); 
    } 

    int nextInt() 
    { 
        return Integer.parseInt(next()); 
    } 

    long nextLong() 
    { 
        return Long.parseLong(next()); 
    } 

    Double nextDouble() 
    { 
        return Double.parseDouble(next());
    } 

    String nextLine() 
    { 
        String str = ""; 
        try
        { 
            str = br.readLine(); 
        } 
        catch (IOException e) 
        { 
            e.printStackTrace(); 
        } 
        return str;
    } 
}
    //====================================================================================================//
