/*

     题意：给出无向无环图，每一个点的度数和相邻点的异或和(a^b^c^....)

     图论/位运算：其实这题很简单。类似拓扑排序，先把度数为1的先入对，每一次少一个度数

                 关键在于更新异或和，精髓：a ^ b = c -> a ^ c = b, b ^ c = a;

 */

 #include <cstdio>

 #include <cstring>

 #include <cmath>

 #include <algorithm>

 #include <queue>

 using namespace std;

 const int MAXN = 7e4 + ;

 const int INF = 0x3f3f3f3f;

 int ans[MAXN][];

 int d[MAXN], s[MAXN];

 int main(void)        //Codeforces Round #285 (Div. 2) C. Misha and Forest

 {

     // freopen ("C.in", "r", stdin);

     int n;

     while (scanf ("%d", &n) == )

     {

         queue<int> Q;

         for (int i=; i<n; ++i)

         {

             scanf ("%d%d", &d[i], &s[i]);

             if (d[i] == )    Q.push (i);

         }

         int cnt = ;

         while (!Q.empty ())

         {

             int u = Q.front ();    Q.pop ();

             if (d[u] == )    continue;

             int v = s[u];

             ans[++cnt][] = u;    ans[cnt][] = v;

             if ((--d[v]) == )    Q.push (v);

             s[v] = s[v] ^ u;

         }

         printf ("%d\n", cnt);

         for (int i=; i<=cnt; ++i)

         {

             printf ("%d %d\n", ans[i][], ans[i][]);

         }

     }

     return ;

 }