#include <bits/stdc++.h>
using namespace std;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
using ll = long long;
using P = pair<ll, ll>;
const long double PI = acos(-1.0L);
ll GCD(ll a, ll b) { return b?GCD(b, a%b):a; }
ll LCM(ll a, ll b) { return a/GCD(a, b)*b; }

// auto mod int
const int mod = 1000000007;
// const int mod = 998244353;
struct mint {
    ll x; // typedef long long ll;
    mint(ll x=0):x((x%mod+mod)%mod){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
      if ((x += a.x) >= mod) x -= mod;
      return *this;
    }
    mint& operator-=(const mint a) {
      if ((x += mod-a.x) >= mod) x -= mod;
      return *this;
    }
    mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this;}
    mint operator+(const mint a) const { return mint(*this) += a;}
    mint operator-(const mint a) const { return mint(*this) -= a;}
    mint operator*(const mint a) const { return mint(*this) *= a;}
    mint pow(ll t) const {
      if (!t) return 1;
      mint a = pow(t>>1);
      a *= a;
      if (t&1) a *= *this;
      return a;
    }

    // for prime mod
    mint inv() const { return pow(mod-2);}
    mint& operator/=(const mint a) { return *this *= a.inv();}
    mint operator/(const mint a) const { return mint(*this) /= a;}
};
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}

int n;
struct Data {
    int no; mint dist;
};

int main() {
    cin >> n;
    vector< vector<int> > Gto(1000100, vector<int>());
    vector< vector<int> > Gfrom(1000100, vector<int>());
    for(int i = 0; i < n-1; ++i) {
        int a, b; cin >> a >> b;
        a--; b--;
        Gto[a].emplace_back(b);
        Gfrom[b].emplace_back(a);
    }

    queue<int> que;
    que.push(0);
    int root = 0;
    while(!que.empty()) {
        int now = que.front(); que.pop();
        // cout << now << endl;
        int glen = Gfrom[now].size();
        if(glen == 0) {
            root = now; break;
        }
        for(int i = 0; i < glen; ++i) {
            que.push(Gfrom[now][i]);
        }
    }

    // cout << root << endl;

    mint ans = 0;
    // 根がどれかはわからないけど木なので正しい根を選べば全部探索できる
    queue<Data> Dque;
    Data first; first.no = root; first.dist = 0;
    Dque.push(first);
    while(!Dque.empty()) {
        Data now = Dque.front(); Dque.pop();
        ans += ((now.dist+1)*now.dist)/2;
        for(int i = 0; i < (int)Gto[now.no].size(); ++i) {
            Data next; next.no = Gto[now.no][i];
            next.dist = now.dist+1;
            Dque.push(next);
        }
    }
    cout << ans << endl;
}