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, *degree*<sub class="lower-index">*v*</sub> and *s*<sub class="lower-index">*v*</sub>, 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 *degree*<sub class="lower-index">*v*</sub> and *s*<sub class="lower-index">*v*</sub> 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.
The first line contains integer *n* (1 ≤ *n* ≤ 2<sup class="upper-index">16</sup>), the number of vertices in the graph.

The *i*-th of the next lines contains numbers *degree*<sub class="lower-index">*i*</sub> and *s*<sub class="lower-index">*i*</sub> (0 ≤ *degree*<sub class="lower-index">*i*</sub> ≤ *n* - 1, 0 ≤ *s*<sub class="lower-index">*i*</sub> &lt; 2<sup class="upper-index">16</sup>), separated by a space.
import sys

def read(T):
	return [T(i) for i in sys.stdin.readline().split()]

def main():
	n=read(int)[0]
	a=[read(int) for i in range(n)]
	stack,res=[],[]
	for i in range(n):
		if a[i][0]==1:
			stack.append(i)

	while len(stack):
		cur=stack.pop()
		if not a[cur][0]:
			continue
		res.append([cur,a[cur][1]])
		a[cur][0]-=1
		a[a[cur][1]][0]-=1
		if a[a[cur][1]][0]==1:
			stack.append(a[cur][1])
		a[a[cur][1]][1]^=cur
		a[cur][1]=0

	print(len(res))
	for edge in res:
		print(edge[0],edge[1])

main()