Heidi has finally found the mythical Tree of Life – a legendary combinatorial structure which is said to contain a prophecy crucially needed to defeat the undead armies.

On the surface, the Tree of Life is just a regular undirected tree well-known from computer science. This means that it is a collection of n points (called vertices), some of which are connected using n - 1 line segments (edges) so that each pair of vertices is connected by a path (a sequence of one or more edges).

To decipher the prophecy, Heidi needs to perform a number of steps. The first is counting the number of lifelines in the tree – these are paths of length 2, i.e., consisting of two edges. Help her!

Input

The first line of the input contains a single integer n – the number of vertices in the tree (1 ≤ n ≤ 10000). The vertices are labeled with the numbers from 1 to n. Then n - 1 lines follow, each describing one edge using two space-separated numbers a b – the labels of the vertices connected by the edge (1 ≤ a < b ≤ n). It is guaranteed that the input represents a tree.

Output

Print one integer – the number of lifelines in the tree.

Examples

Input

4
1 2
1 3
1 4


Output

3

Input

5
1 2
2 3
3 4
3 5


Output

4

Note

In the second sample, there are four lifelines: paths between vertices 1 and 3, 2 and 4, 2 and 5, and 4 and 5.
#include <bits/stdc++.h>
struct graph {
  int v;
  struct graph *next;
} * remember[11000], *arr[11000];
int visited[11000];
int depth[11000];
int outDEG[11000];
void insert_edge(int v, int v1) {
  struct graph *temp = (struct graph *)malloc(sizeof(struct graph));
  temp->next = NULL;
  temp->v = v1;
  if (!arr[v]) {
    arr[v] = temp;
    remember[v] = temp;
    return;
  }
  remember[v]->next = temp;
  remember[v] = temp;
}
void dfs(int v, int parent, int far) {
  if (visited[v]) return;
  visited[v] = 1;
  if (parent != -1) {
    outDEG[parent]++;
  }
  depth[v] = far;
  struct graph *sweep = arr[v];
  while (sweep) {
    dfs(sweep->v, v, far + 1);
    sweep = sweep->next;
  }
}
int solve(int x) { return (x * (x + 1)) / 2; }
int main() {
  int a, b, c, d, e;
  scanf("%d", &a);
  int i;
  for (i = 0; i < a - 1; i++) {
    int in, in1;
    scanf("%d%d", &in, &in1);
    insert_edge(in, in1);
    insert_edge(in1, in);
  }
  dfs(1, -1, 1);
  int cnt = 0;
  for (i = 1; i <= a; i++) {
    if (depth[i] >= 3) cnt++;
    if (outDEG[i] > 1) {
      cnt = cnt + solve(outDEG[i] - 1);
    }
  }
  printf("%d", cnt);
}
