Codeforces Round #475 (Div. 1) B. Destruction of a Tree

B. Destruction of a Tree

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a tree (a graph with n vertices and n - 1 edges in which it's possible to reach any vertex from any other vertex using only its edges).

A vertex can be destroyed if this vertex has even degree. If you destroy a vertex, all edges connected to it are also deleted.

Destroy all vertices in the given tree or determine that it is impossible.

Input

The first line contains integer n (1 ≤ n ≤ 2·105) — number of vertices in a tree.

The second line contains n integers p1, p2, ..., pn (0 ≤ pi ≤ n). If pi ≠ 0 there is an edge between vertices i and pi. It is guaranteed that the given graph is a tree.

Output

If it's possible to destroy all vertices, print "YES" (without quotes), otherwise print "NO" (without quotes).

If it's possible to destroy all vertices, in the next n lines print the indices of the vertices in order you destroy them. If there are multiple correct answers, print any.

Examples

input

Copy

5
0 1 2 1 2

output

Copy

YES
1
2
3
5
4

input

Copy

4
0 1 2 3

output

Copy

NO

思路：贪心消除最靠近叶子的节点。因为如果最靠近叶子的偶数度节点晚于父节点消除，则父节点消除后此节点度数变为奇数，且其所有子节点度数都为奇数，就再也消除不了了。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
*/

int n, p[200015], vis[200015], deg[200015], par[200015];
vector<int> edge[200015];
stack<int> sta;
vector<int> ans;
void dfs2(int x) {
    ans.pb(x);
    vis[x] = 1;
    for (int i = 0; i < edge[x].size(); ++i) {
        int y = edge[x][i];
        --deg[y];
        if (y == par[x]) continue;
        if (vis[y]) continue;
        if (deg[y] % 2 == 0) {
            dfs2(y);
        }
    }
}
void dfs(int p, int x) {
    sta.push(x);
    par[x] = p;
    for (int i = 0; i < edge[x].size(); ++i) {
        int y = edge[x][i];
        if (y == p) continue;
        dfs(x, y);
    }
}
int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; ++i) {
        scanf("%d", &p[i]);
        if (p[i]) {
            edge[p[i]].pb(i);
            edge[i].pb(p[i]);
            ++deg[p[i]];
            ++deg[i];
        }
    }
    dfs(0, 1);
    while (sta.size()) {
        int x = sta.top(); sta.pop();
        if (deg[x] % 2 == 0) {
            dfs2(x);
        }
    }
    if (ans.size() == n) {
        puts("YES");
        for (int i = 0; i < ans.size(); ++i) {
            printf("%d
", ans[i]);
        }
    } else {
        puts("NO");
    }
    return 0;
}