題干：

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".

題目大意：

有n個點，代號分別為0到n-1,然后第i個點有di個相連點，與i 相連的點的編號的XOR sum 為si，求所有的邊。

輸入n，然后n行，每行di和si。

解題報告：

? ? 因為保證題目有解，那也就是一定有葉子結點唄，，度數為1，直接拓撲不就完了。注意如果度數已經為0了要continue掉，因為這時候保證一定不會有和他相連的邊了，，但是不能si==0就continue掉啊，，因為雖然a^a=0，并且0~n-1不會有重復的值，但是說不定幾個數的異或值就會出現a^b^c^....^z=0這樣的情況呢、、、

AC代碼：

#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 ;}

?

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