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.

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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:
/*mjytxdy!*/
#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = (a); i <= int(b); i++)
#define per(i, a, b) for(int i = (a); i >= int(b); i--)
#define fi first
#define se second
#define all(s) s.begin(),s.end()
#define trv(i, u) for(int i = head[u]; i; i = e[i].next)
#define sz(s) (int)(s.size())
using namespace std;

typedef long long ll;
typedef pair<int,int> P;

template <typename T> bool chkmx(T &x, T y){return x < y ? x = y, true : false;}
template <typename T> bool chkmn(T &x, T y){return x > y ? x = y, true : false;}

const int maxn = 1e5;
const int B =  500;
const int inf = 1e9;

int n, x, u, mxdep, dep[maxn + 5], k[maxn + 5], col[maxn + 5];
bitset<maxn + 5> dp[B + 5];
int pre[B + 5][maxn + 5], cnt[maxn + 5];
vector<int> G[maxn + 5], p[maxn + 5], s;
vector<P> t;

void dfs(int u, int par) {
	p[dep[u] = dep[par] + 1].push_back(u);
	chkmx(mxdep, dep[u]); 
	for(auto v : G[u]) dfs(v, u);
}
int main() {
//	freopen("in.txt", "r", stdin);
	scanf("%d%d", &n, &x);
	rep(i, 2, n) {
		scanf("%d", &u);
		G[u].push_back(i);
	}
	dfs(1, 1);
	rep(i, 1, mxdep) s.push_back(sz(p[i]));
	sort(all(s));
	rep(i, 0, sz(s) - 1) {
		if(i == 0 || s[i] != s[i - 1]) t.push_back({s[i], 1});
		else t.back().se++;
	}
	rep(i, 0, sz(t) - 1) {
		if(i) rep(j, 0, x) dp[i][j] = dp[i - 1][j];
		else dp[i][0] = 1;
		rep(j, 0, x) {
			if(dp[i][j]) k[j] = 0;
			else k[j] = j >= t[i].fi ? k[j - t[i].fi] + 1 : inf;
			if(k[j] <= t[i].se) dp[i][j] = 1, pre[i][j] = j - t[i].fi * k[j];
			else dp[i][j] = 0;
		}
	}
	if(dp[sz(t) - 1][x]) {
		int nw = x;
		per(i, sz(t) - 1, 0) {
			cnt[t[i].fi] = (nw - pre[i][nw]) / t[i].fi;
			nw = pre[i][nw];
		}
		rep(i, 1, mxdep) {
			if(cnt[sz(p[i])]) {
				cnt[sz(p[i])]--;
				for(auto u : p[i]) col[u] = 1;
			}
			else for(auto u : p[i]) col[u] = 2;
		}
		printf("%d\n", mxdep);
		rep(i, 1, n) assert(col[i]), putchar(col[i] == 1 ? 'a' : 'b');
	}
	else {
		int m = n, y = x;
		rep(i, 1, mxdep) {
			if(sz(p[i]) <= y) for(auto u : p[i]) col[u] = 1, y--, m--;
			else if(sz(p[i]) <= m - y) for(auto u : p[i]) col[u] = 2, m--;
			else {
				if(y < m - y) {
					for(auto u : p[i]) if(!G[u].empty()) col[u] = 2, m--;
					for(auto u : p[i]) if(G[u].empty()) col[u] = (m - y) ? (m--, 2) : (m--, y--, 1);
				}
				else {
					for(auto u : p[i]) if(!G[u].empty()) col[u] = 1, y--, m--;
					for(auto u : p[i]) if(G[u].empty()) col[u] = (y) ? (m--, y--, 1) : (m--, 2);
				}
			}
		}
		printf("%d\n", mxdep + 1);
		rep(i, 1, n) assert(col[i]), putchar(col[i] == 1 ? 'a' : 'b');
	}
	return 0;
}