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:
#include <bits/stdc++.h>
#define IO_OP std::ios::sync_with_stdio(0); std::cin.tie(0);
#define F first
#define S second
#define V vector
#define PB push_back
#define MP make_pair
#define EB emplace_back
#define ALL(v) (v).begin(), (v).end()
#define debug(x) cerr << "Line(" << __LINE__ << ") -> " << #x << " is " << x << endl

using namespace std;

typedef long long ll;
typedef pair<int, int> pi;
typedef V<int> vi;

const int INF = 1e9 + 7, N = 1e5 + 7;

vi G[N], who[N];
int cnt[N], cnt2[N], mx_depth;

void dfs(int u, int depth = 0) {
	mx_depth = max(mx_depth, depth);
	cnt[depth]++;
	who[depth].PB(u);
	for(int v:G[u]) dfs(v, depth + 1);
}

signed main()
{
	IO_OP;

	int n, x;
	cin >> n >> x;
	for(int i = 2; i <= n; i++) {
		int p;
		cin >> p;
		G[p].PB(i);
	}
	dfs(1);
	for(int i = 0; i < n; i++)
		if(cnt[i])
			cnt2[cnt[i]]++;
	vi dp(n + 1);
	V<pi> from(n + 1);
	dp[0] = 1;
	for(int i = 1; i <= n; i++) if(cnt2[i]) { // this if is true for at most sqrt(n) times
		vi new_dp = dp;
		vi lst(i, -1);
		for(int j = 0; j <= n; j++) {
			int k = j % i;
			if(lst[k] != -1 && (j - lst[k]) / i <= cnt2[i]) {
				new_dp[j] = 1;
				if(from[j] == MP(0, 0))
					from[j] = {i, (j - lst[k]) / i};
			}
			if(dp[j]) lst[k] = j;
		}
		swap(dp, new_dp);
	}
	if(dp[x]) {
		vi need(n + 1);
		while(x) {
			pi p = from[x];
			x -= p.F * p.S;
			need[p.F] = p.S;
		}
		string ans(n, 'b');
		for(int i = 0; i < n; i++) if(need[cnt[i]]) {
			need[cnt[i]]--;
			for(int u:who[i])
				ans[u - 1] = 'a';
		}
		cout << mx_depth + 1 << '\n';
		cout << ans << '\n';
	} else {
		int a = x, b = n - x;
		string ans(n, 'c');
		for(int i = 0; i < n; i++) {
			if(who[i].empty()) break;
			if(who[i].size() <= a) {
				for(int u:who[i])
					ans[u - 1] = 'a', a--;
			} else if(who[i].size() <= b) {
				for(int u:who[i])
					ans[u - 1] = 'b', b--;
			} else {
				for(int u:who[i]) if(G[u].empty()) {
					if(min(a, b) == 0) break;
					if(a <= b)
						ans[u - 1] = 'a', a--;
					else
						ans[u - 1] = 'b', b--;
				}
				assert(a == 0 || b == 0);
				for(int u:who[i]) if(ans[u - 1] == 'c') {
					if(a) ans[u - 1] = 'a', a--;
					else if(b) ans[u - 1] = 'b', b--;
				}
			}
		}
		cout << mx_depth + 2 << '\n';
		cout << ans << '\n';
	}
}