Let’s define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n vertices. For each vertex v from 0 to n - 1 he wrote down two integers, degreev and sv, were the first integer is the number of vertices adjacent to vertex v, and the second integer is the XOR sum of the numbers of vertices adjacent to v (if there were no adjacent vertices, he wrote down 0).
Next day Misha couldn’t remember what graph he initially had. Misha has values degreev and sv left, though. Help him find the number of edges and the edges of the initial graph. It is guaranteed that there exists a forest that corresponds to the numbers written by Misha.
Input
The first line contains integer n (1 ≤ n ≤ 216), the number of vertices in the graph.
The i-th of the next lines contains numbers degreei and si (0 ≤ degreei ≤ n - 1, 0 ≤ si < 216), separated by a space.
Output
In the first line print number m, the number of edges of the graph.
Next print m lines, each containing two distinct numbers, a and b (0 ≤ a ≤ n - 1, 0 ≤ b ≤ n - 1), corresponding to edge (a, b).
Edges can be printed in any order; vertices of the edge can also be printed in any order.
Examples
inputCopy
3
2 3
1 0
1 0
outputCopy
2
1 0
2 0
inputCopy
2
1 1
1 0
outputCopy
1
0 1
Note
The XOR sum of numbers is the result of bitwise adding numbers modulo 2. This operation exists in many modern programming languages. For example, in languages C++, Java and Python it is represented as “^”, and in Pascal — as “xor”.
给定节点的度数以及与其相邻节点的异或和，求边数以及边的两个端点；
思路：
从叶子入手，其异或和就是其父节点的编号，用 队列进行维护一下，记得到叶子节点时更新其父节点时degree–；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<string>
#include<cstring>
#include<set>
#include<queue>
#include<map>
using namespace std;
#define maxn 100005
#define inf 0x3f3f3f3f
typedef long long ll;
typedef unsigned long long int ull;
struct node
{
int degree,s,i;
int vis;
void output()
{
cout<<i<<' '<<degree<<' '<<s<<endl;
}
}v[maxn];
typedef pair<int,int> pii;
vector<pii> edge;
vector<pii>::iterator it;
queue<int>q;
int main()
{
ios::sync_with_stdio(false);
int n;cin>>n;
int cnt=0;
for(int i=0;i<n;i++){
cin>>v[i].degree>>v[i].s;
if(v[i].degree==1)q.push(i);
}
while(!q.empty()){
int i=q.front();
q.pop();
if(v[i].degree==1){
int fath=v[i].s;
v[fath].s=i^v[fath].s;
v[fath].degree--;
edge.push_back(make_pair(i,fath));
if(v[fath].degree==1)q.push(fath);
}
}
cout<<edge.size()<<endl;
for(it=edge.begin();it!=edge.end();it++){
pii p=*it;
cout<<p.first<<' '<<p.second<<endl;
}
}