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>
using namespace std;
const int maxn = 1e4 + 10;
int m, n, vis[maxn];
int fa[maxn], son[maxn];
vector<int> g[maxn];
int ans;
void init() {
  for (int i = 1; i <= n; i++) {
    g[i].clear();
    vis[i] = 0;
    fa[i] = 0;
    son[i] = 0;
    fa[i] = 1;
  }
}
void dfs(int u) {
  vis[u] = 1;
  int cnt = 0;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (!vis[v]) {
      int t = son[u] - 1;
      if (!cnt) ans += (t * t + t) / 2;
      ans += fa[v] * son[v];
      cnt++;
      dfs(v);
    }
  }
}
void dfss(int u) {
  vis[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (!vis[v]) {
      son[u]++;
      dfss(v);
    }
  }
}
int main() {
  while (~scanf("%d", &n)) {
    int cnt = 0;
    int a, b;
    init();
    for (int i = 0; i < n - 1; i++) {
      scanf("%d%d", &a, &b);
      g[a].push_back(b);
      g[b].push_back(a);
    }
    ans = 0;
    dfss(1);
    fa[1] = 0;
    memset(vis, 0, sizeof(vis));
    dfs(1);
    printf("%d\n", ans);
  }
  return 0;
}
