Problem: Kilani and Abd are neighbors for 3000 years, but then the day came and Kilani decided to move to another house. As a farewell gift, Kilani is going to challenge Abd with a problem written by their other neighbor with the same name Abd.

<image>

The problem is:

You are given a connected tree rooted at node 1.

You should assign a character a or b to every node in the tree so that the total number of a's is equal to x and the total number of b's is equal to n - x.

Let's define a string for each node v of the tree as follows: 

  * if v is root then the string is just one character assigned to v: 
  * otherwise, let's take a string defined for the v's parent p_v and add to the end of it a character assigned to v. 



You should assign every node a character in a way that minimizes the number of distinct strings among the strings of all nodes.

Input

The first line contains two integers n and x (1 ≤ n ≤ 10^5; 0 ≤ x ≤ n) — the number of vertices in the tree the number of a's.

The second line contains n - 1 integers p_2, p_3, ..., p_{n} (1 ≤ p_i ≤ n; p_i ≠ i), where p_i is the parent of node i.

It is guaranteed that the input describes a connected tree.

Output

In the first line, print the minimum possible total number of distinct strings.

In the second line, print n characters, where all characters are either a or b and the i-th character is the character assigned to the i-th node.

Make sure that the total number of a's is equal to x and the total number of b's is equal to n - x.

If there is more than one answer you can print any of them.

Example

Input


9 3
1 2 2 4 4 4 3 1


Output


4
aabbbbbba

Note

The tree from the sample is shown below:

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The tree after assigning characters to every node (according to the output) is the following:

<image>

Strings for all nodes are the following: 

  * string of node 1 is: a
  * string of node 2 is: aa
  * string of node 3 is: aab
  * string of node 4 is: aab
  * string of node 5 is: aabb
  * string of node 6 is: aabb
  * string of node 7 is: aabb
  * string of node 8 is: aabb
  * string of node 9 is: aa



The set of unique strings is \{a, aa, aab, aabb\}, so the number of distinct strings is 4.

Solution (in CPP):

Solution:
/*CF1481F AB Tree*/
#include<bits/stdc++.h>
using namespace std;
#define ll long long
int read(){
	char c = getchar();
	int x = 0;
	while(c < '0' || c > '9')		c = getchar();
	while(c >= '0' && c <= '9')		x = x * 10 + c - 48,c = getchar();
	return x;
}
const int N = 1e5 + 10; 
bool f[N],g[N];int n,x;
typedef pair<int,int> pii;
#define mp make_pair
#define fi first
#define se second
pii pf[N];int pg[N];
int cdep[N],c[N];/*记录深度为d的有多少个*/
int dep[N];int is_A[N];
struct Edge{
	int nxt,point;
}edge[N<<1];int head[N],tot;
void add_edge(int u,int v){
	edge[++tot].nxt = head[u];
	edge[tot].point = v;
	head[u] = tot;
}
void dfs(int u){
	cdep[dep[u]]++;
	for(int i = head[u]; i ; i = edge[i].nxt){
		int v = edge[i].point;
		dep[v] = dep[u] + 1;
		dfs(v);
	}
}
int st[N],top,mxdep = 0;
vector<int>Dep[N];
#define pb push_back 
void build(int s,int t){/*f数组花了s,g数组花了t*/
	//is_A数组记录答案中大小为d的物品被用了多少次 
	while(s){
		pii now = pf[s];
		int fi = now.fi,se = now.se;
		is_A[fi] += se;
		s -= fi * se; 
	}
	while(t){
		int x = pg[t];
		is_A[x]++;
		t -= x;
	} 
	static bool flag[N];memset(flag,0,sizeof flag);
	for(int d = 1; d <= n; ++d){
		if(is_A[cdep[d]] != 0){
			flag[d] = 1;
			is_A[cdep[d]]--;
		}
	}
	for(int i = 1; i <= n; ++i){
		if(flag[dep[i]] == 1)	putchar('a');
		else	putchar('b');
	}
	return;
}
bool check(){
	int Sqrn = sqrt(n);
	f[0] = 1;
	for(int i = 1; i <= n; ++i)		mxdep = max(mxdep,dep[i]);
	for(int d = 1; d <= n; ++d)		c[cdep[d]]++;/*相当于记录大小为cdep[d]的物品有多少*/
	for(int d = 1; d <= Sqrn; ++d){/*该物品大小为d*/
		int x = c[d];
		top = 0;int base = 1;
		for(base = 1; base <= x; base <<= 1) {
			st[++top] = base * d;
			x -= base;
		}
		st[++top] = x * d;
		for(int k = 1; k <= top; ++k){
			for(int i = n; i >= st[k]; --i){
				if(f[i])	continue;
				if(f[i - st[k]]){
					f[i] = 1;
					pf[i] = mp(d,st[k]/d);/*从se个大小为d的层转移来的*/
				}
			}
		}
	}
	g[0] = 1;
	for(int d = Sqrn + 1; d <= n; ++d){
		int x = c[d];if(!x)		continue;
		for(int k = 1; k <= x; ++k){
			for(int i = n; i >= d; --i){
				if(g[i])	continue;
				if(g[i - d]){
					g[i] = 1;
					pg[i] = d;
				}
			}
		}
	}
	for(int i = 0; i <= x; ++i){
		if(f[i] && g[x-i]){
			printf("%d\n",mxdep);
			build(i,x-i);
			return 1;
		}
	}
	return 0; 
}
void solve(int s,int t){
	memset(is_A,0,sizeof is_A);
	for(int i = 1; i <= n; ++i)		Dep[dep[i]].pb(i);
	for(int d = 1; d <= n; ++d){
		if(!s && !t)	break;
		if(!s){
			continue;
		}
		if(!t){
			for(int i = 0; i < (int)Dep[d].size(); ++i){
				is_A[Dep[d][i]] = 1;
			}
			continue;
		}
		if(cdep[d] <= s && cdep[d] <= t){
			t -= cdep[d];
		}
		else{
			if(cdep[d] <= s){
				s -= cdep[d];
				for(int i = 0; i < (int)Dep[d].size(); ++i){
					int u = Dep[d][i];
					is_A[u] = 1;
				}
			}
			else if(cdep[d] <= t){
				t -= cdep[d];
			}
			else{
				for(int i = 0; i < (int)Dep[d].size(); ++i)		is_A[Dep[d][i]] = -1;
				if(s < t){
					for(int i = 0; i < (int)Dep[d].size(); ++i){
						if(head[Dep[d][i]])		is_A[Dep[d][i]] = 0,t--;
					}
					for(int i = 0; i < (int)Dep[d].size(); ++i){
						if(is_A[Dep[d][i]] == -1 && t){
							is_A[Dep[d][i]] = 0;t--;
						}
					}
					for(int i = 0; i < (int)Dep[d].size(); ++i){
						if(is_A[Dep[d][i]] == -1)		is_A[Dep[d][i]] = 1;
					}
				}
				else{
					for(int i = 0; i < (int)Dep[d].size(); ++i){
						if(head[Dep[d][i]])		is_A[Dep[d][i]] = 1,s--;
					}
					for(int i = 0; i < (int)Dep[d].size(); ++i){
						if(is_A[Dep[d][i]] == -1 && s){
							is_A[Dep[d][i]] = 1;s--;
						}
					}
					assert(s == 0);
					for(int i = 0; i < (int)Dep[d].size(); ++i){
						if(is_A[Dep[d][i]] == -1)		is_A[Dep[d][i]] = 0;					
					}
				}
			}
		}
	}
	for(int i = 1; i <= n; ++i){
		if(is_A[i])		putchar('a');
		else	putchar('b');
	}
}
int main(){
	n = read(),x = read();
	for(int i = 2; i <= n; ++i){
		int fa = read();
		add_edge(fa,i); 
	}
	dep[1] = 1;
	dfs(1);
	if(check()){
		return 0;
	}
	cout<<mxdep+1<<endl;
	solve(x,n-x);
	return 0;
}