题干：

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".

题目大意：

有n个点，代号分别为0到n-1,然后第i个点有di个相连点，与i 相连的点的编号的XOR sum 为si，求所有的边。

输入n，然后n行，每行di和si。

解题报告：

    因为保证题目有解，那也就是一定有叶子结点呗，，度数为1，直接拓扑不就完了。注意如果度数已经为0了要continue掉，因为这时候保证一定不会有和他相连的边了，，但是不能si==0就continue掉啊，，因为虽然a^a=0，并且0~n-1不会有重复的值，但是说不定几个数的异或值就会出现a^b^c^....^z=0这样的情况呢、、、

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int d[MAX],sum[MAX],ans;
int main()
{
	queue<int> q;
	int n;
	cin>>n;
	for(int i = 0; i<n; i++) {
		scanf("%d%d",d+i,sum+i);
		if(d[i] == 1) q.push(i);
		ans += d[i];
	}
	printf("%d\n",ans/2);
	while(!q.empty()) {
		int cur = q.front();q.pop();
		if(d[cur]==0) continue;  
		int x1 = cur;
		int x2 = sum[cur];
		sum[x2]^=x1;
		d[x2]--;
		printf("%d %d\n",x1,x2);
		if(d[x2]==1) q.push(x2);
	}
	return 0 ;
 }