CF982C Cut 'em all! DFS 树 * 二十一

 Cut 'em all!

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You're given a tree with n vertices.

Your task is to determine the maximum possible number of edges that can be removed in such a way that all the remaining connected components will have even size.

Input

The first line contains an integer 1≤n≤105) denoting the size of the tree.

The next i-th edge.

It's guaranteed that the given edges form a tree.

Output

Output a single integer −1 if it is impossible to remove edges in order to satisfy this property.

Examples

input

Copy

4
2 4
4 1
3 1

output

Copy

1

input

Copy

3
1 2
1 3

output

Copy

-1

input

Copy

10
7 1
8 4
8 10
4 7
6 5
9 3
3 5
2 10
2 5

output

Copy

4

input

Copy

2
1 2

output

Copy

0

Note

In the first example you can remove the edge between vertices 4. The graph after that will have two connected components with two vertices in each.

In the second example you can't remove edges in such a way that all components have even number of vertices, so the answer is −1.

 

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
int hd[maxn], ne[maxn*2], to[maxn*2], num, n, siz[maxn], ans;
void add( int x, int y ) {
    to[++num]=y, ne[num]=hd[x], hd[x]=num;
}
 
void dfs( int x, int fa ) {
    siz[x]=1;
    for( int i = hd[x]; i; i = ne[i] ) {
        if(to[i]!=fa){
            dfs(to[i],x);
            siz[x]+=siz[to[i]];
        }
    }
    if(!(siz[x]&1)) 
        siz[x]=0,ans++;
}
 
int main(){
    std::ios::sync_with_stdio(false);
    scanf("%d",&n);
    int uu, vv;
    for( int i = 1; i < n; i ++ ) {
        scanf("%d%d",&uu,&vv);
        add(uu,vv), add(vv,uu);
    }
    if( n & 1 ) { 
        puts("-1"); 
        return 0;
    }
     
    dfs( 1, -1 ), ans--;
     
    printf("%d\n", ans );
    return 0;
}