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.
R = lambda: map(int, input().split())
n = int(input())
degs, xors = [0] * (2 ** 16 + 1), [0] * (2 ** 16 + 1)
edges = []
for curr in range(n):
    degs[curr], xors[curr] = R()
q = []
for curr in range(n):
    if degs[curr] == 1:
        q.append(curr)
while q:
    curr = q.pop(0)
    if degs[curr] != 1:
        continue
    neighbor = xors[curr]
    edges.append(sorted([curr, neighbor]))
    degs[neighbor] -= 1
    xors[neighbor] ^= curr
    if degs[neighbor] == 1:
        q.append(neighbor)
filter(lambda p: p[0] < p[1], edges)
print(len(edges))
for u, v in edges:
    if u < v:
        print(u, v)