CF1541

  1 #include <bits/stdc++.h>
  2 
  3 typedef long long LL;
  4 
  5 const int N = 210;
  6 const LL MO = 1e9 + 7, inv2 = 5e8 + 4;
  7 
  8 struct Edge {
  9     int nex, v;
 10 }edge[N * 2]; int tp;
 11 
 12 int n, e[N], siz[N], d[N], fa[N];
 13 LL f[N][N], g[N][N];
 14 LL ans;
 15 
 16 LL qpow(LL a, LL b) {
 17     a %= MO;
 18     LL ans = 1;
 19     while(b) {
 20         if(b & 1) {
 21             ans = ans * a % MO;
 22         }
 23         a = a * a % MO;
 24         b = b >> 1;
 25     }
 26     return ans;
 27 }
 28 
 29 inline void add(int x, int y) {
 30     ++tp;
 31     edge[tp].v = y;
 32     edge[tp].nex = e[x];
 33     e[x] = tp;
 34     return;
 35 }
 36 
 37 void DFS1(int x, int f) {
 38     siz[x] = 1;
 39     d[x] = d[f] + 1;
 40     fa[x] = f;
 41     for(int i = e[x]; i; i = edge[i].nex) {
 42         int y = edge[i].v;
 43         if(y == f) {
 44             continue;
 45         }
 46         DFS1(y, x);
 47         siz[x] += siz[y];
 48     }
 49     return;
 50 }
 51 
 52 void DFS2(int x, int r) {
 53     for(int i = e[x]; i; i = edge[i].nex) {
 54         int y = edge[i].v;
 55         if(y == fa[x]) {
 56             continue;
 57         }
 58         DFS2(y, r);
 59     }
 60     if(x > r) {
 61         int p = x;
 62         while(fa[p] != r) {
 63             // cal fa[p]
 64             ans += 1ll * (siz[fa[p]] - siz[p]) * g[d[fa[p]]][d[x] - d[fa[p]]] % MO;
 65             ans %= MO;
 66             // printf("ans += %d * %lld \n", siz[fa[p]] - siz[p], g[d[fa[p]]][d[x] - d[fa[p]]]);
 67             p = fa[p];
 68         }
 69         ans = (ans + siz[x]) % MO;
 70         // printf("ans += %d \n", siz[x]);
 71     }
 72     return;
 73 }
 74 
 75 int main() {
 76     scanf("%d", &n);
 77     for(int i = 1, x, y; i < n; i++) {
 78         scanf("%d%d", &x, &y);
 79         add(x, y);
 80         add(y, x);
 81     }
 82     
 83     // prework
 84     f[0][0] = 1;
 85     for(int i = 1; i <= n; i++) {
 86         f[i][0] = f[0][i] = f[0][i - 1] * inv2 % MO;
 87     }
 88     for(int i = 1; i <= n; i++) {
 89         for(int j = 1; j <= n; j++) {
 90             f[i][j] = f[i][j - 1] * inv2 + f[i - 1][j] * inv2;
 91             f[i][j] %= MO;
 92         }
 93     }
 94     for(int i = 0; i <= n; i++) {
 95         for(int j = 1; j <= n; j++) {
 96             // cal g[i][j]
 97             for(int k = 0; k < i; k++) {
 98                 g[i][j] = (g[i][j] + f[k][j - 1] * inv2 % MO) % MO;
 99             }
100         }
101     }
102 
103     // printf("%lld \n", g[2][1]);
104 
105     // solve
106     d[0] = -1;
107     for(int x = 1; x <= n; x++) {
108         // x is smaller one
109         DFS1(x, 0);
110         DFS2(x, x);
111         // puts("");
112     }
113     // printf("ans = %lld\n", ans);
114     printf("%lld\n", ans * qpow(n, MO - 2) % MO);
115     return 0;
116 }