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
Input
3 2 3 1 0 1 0
Output
2 1 0 2 0
Input
2 1 1 1 0
Output
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".
题意：给出0~n-1个点的度，以及和这个点相邻的所有点的异或值，问原始的树的边数和每条边的端点。
题解：拓扑一边就可以了，度数为1的点的价值即为连接的其他点的编号，注意每个点只操作一次。
#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
const int N=1e6+10;
const ll mod=1e9+7;
vector<int> v[N];
struct node{
int id;
int du,val;
}a[N];
int n;
int main()
{
while(~scanf("%d",&n))
{
queue<int> q;
int cn=0;
for(int i=0;i<n;i++)
{
scanf("%d%d",&a[i].du,&a[i].val);
a[i].id=i;
if(a[i].du==1)
q.push(i);
v[i].clear();
}
while(!q.empty())
{
int now=q.front();q.pop();
if(a[now].du<1) continue;
cn++;
int to=a[now].val;
a[now].du--;
v[now].push_back(a[now].val);
a[to].du--;
a[to].val=a[to].val^now;
if(a[to].du==1) q.push(to);
}
printf("%d\n",cn);
for(int i=0;i<n;i++)
{
for(int j=0;j<v[i].size();j++)
printf("%d %d\n",i,v[i][j]);
}
}
return 0;
}