Problem: Ivan wants to write a letter to his friend. The letter is a string s consisting of lowercase Latin letters.

Unfortunately, when Ivan started writing the letter, he realised that it is very long and writing the whole letter may take extremely long time. So he wants to write the compressed version of string s instead of the string itself.

The compressed version of string s is a sequence of strings c1, s1, c2, s2, ..., ck, sk, where ci is the decimal representation of number ai (without any leading zeroes) and si is some string consisting of lowercase Latin letters. If Ivan writes string s1 exactly a1 times, then string s2 exactly a2 times, and so on, the result will be string s.

The length of a compressed version is |c1| + |s1| + |c2| + |s2|... |ck| + |sk|. Among all compressed versions Ivan wants to choose a version such that its length is minimum possible. Help Ivan to determine minimum possible length.

Input

The only line of input contains one string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 8000).

Output

Output one integer number — the minimum possible length of a compressed version of s.

Examples

Input

aaaaaaaaaa


Output

3


Input

abcab


Output

6


Input

cczabababab


Output

7

Note

In the first example Ivan will choose this compressed version: c1 is 10, s1 is a.

In the second example Ivan will choose this compressed version: c1 is 1, s1 is abcab.

In the third example Ivan will choose this compressed version: c1 is 2, s1 is c, c2 is 1, s2 is z, c3 is 4, s3 is ab.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class F implements Runnable{
	public static void main (String[] args) {new Thread(null, new F(), "_cf", 1 << 28).start();}

	public void run() {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		System.err.println("Go!");

		//aacaacaacaac
		//X10123456789
		//abcdab
		//X00012
		//freaky.
		
		int[] vals = new int[10000];
		for(int i = 1; i < vals.length; i++) {
			int ret = 0, j = i;
			while(j > 0) {
				ret++; j /= 10;
			}
			vals[i] = ret;
		}
		
		String s = fs.next();
		int n = s.length(), oo = (int)2e9;
		int[] dp = new int[n + 1];
		for(int i = n-1; i >= 0; i--) {
			dp[i] = oo;
			int[] kmp = KMP_Construct(s.substring(i).toCharArray());
			for(int j = i, len = 1; j < n; j++, len++) {
				int p = len;
				if(len % (len - kmp[len]) == 0) p = len - kmp[len];
				int cnt = len / p;
				dp[i] = Math.min(dp[i], vals[cnt] + p + dp[j+1]);
			}
		}
		out.println(dp[0]);
		out.close();
	}

	// Given a pattern string w, return the prefix array. //%
	// p[i] indicates the length of the longest prefix //%
	// that is also a substring ending at i.  O(|w|) //%
	static int[] KMP_Construct(char[] w) {
		int n = w.length + 1, k = 0;  int[] p = new int[n];  p[1] = 0;
		for(int i = 2; i < n; ++i) {
			while(k > 0 && w[k] != w[i-1]) k = p[k];
			if(w[k] == w[i-1]) k++;
			p[i] = k;
		}
		return p;
	}

	// Given the prefix array of a pattern, search the //%
	// text for all occurrences of the pattern.  O(|s|) //%
	static ArrayList<Integer> search(char[] s, char[] w, int[] p) {
		int k = 0;  ArrayList<Integer> locs = new ArrayList<>();
		for(int i = 0; i < s.length; ++i) {
			while(k > 0 && w[k] != s[i]) k = p[k];
			if(w[k] == s[i]) k++;
			if(k == w.length) { locs.add(i);  k = p[k]; }
		}
		return locs;
	}

	void sort (int[] a) {
		int n = a.length;
		for(int i = 0; i < 50; i++) {
			Random r = new Random();
			int x = r.nextInt(n), y = r.nextInt(n);
			int temp = a[x];
			a[x] = a[y];
			a[y] = temp;
		}
		Arrays.sort(a);
	}

	class FastScanner {
		public int BS = 1<<16;
		public char NC = (char)0;
		byte[] buf = new byte[BS];
		int bId = 0, size = 0;
		char c = NC;
		double num = 1;
		BufferedInputStream in;

		public FastScanner() {
			in = new BufferedInputStream(System.in, BS);
		}

		public FastScanner(String s) throws FileNotFoundException {
			in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
		}

		public char nextChar(){
			while(bId==size) {
				try {
					size = in.read(buf);
				}catch(Exception e) {
					return NC;
				}                
				if(size==-1)return NC;
				bId=0;
			}
			return (char)buf[bId++];
		}

		public int nextInt() {
			return (int)nextLong();
		}

		public long nextLong() {
			num=1;
			boolean neg = false;
			if(c==NC)c=nextChar();
			for(;(c<'0' || c>'9'); c = nextChar()) {
				if(c=='-')neg=true;
			}
			long res = 0;
			for(; c>='0' && c <='9'; c=nextChar()) {
				res = (res<<3)+(res<<1)+c-'0';
				num*=10;
			}
			return neg?-res:res;
		}

		public double nextDouble() {
			double cur = nextLong();
			return c!='.' ? cur:cur+nextLong()/num;
		}

		public String next() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c>32) {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public String nextLine() {
			StringBuilder res = new StringBuilder();
			while(c<=32)c=nextChar();
			while(c!='\n') {
				res.append(c);
				c=nextChar();
			}
			return res.toString();
		}

		public boolean hasNext() {
			if(c>32)return true;
			while(true) {
				c=nextChar();
				if(c==NC)return false;
				else if(c>32)return true;
			}
		}
		
		public int[] nextIntArray(int n) {
			int[] res = new int[n];
			for(int i = 0; i < n; i++) res[i] = nextInt();
			return res;
		}
		
	}

}