Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices numbered 1 to N.
The i-th edge in this tree connects Vertex a_i and Vertex b_i.

Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?

The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.

Constraints

2 \leq N \leq 50

1 \leq a_i,b_i \leq N

The graph given in input is a tree.

1 \leq M \leq \min(20,\frac{N(N-1)}{2})

1 \leq u_i < v_i \leq N

If i \not= j, either u_i \not=u_j or v_i\not=v_j

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M

Output

Print the number of ways to paint the edges that satisfy all of the M conditions.

Sample Input 1

3
1 2
2 3
1
1 3

Sample Output 1

3

The tree in this input is shown below:

All of the M restrictions will be satisfied if Edge 1 and 2 are respectively painted (white, black), (black, white), or (black, black), so the answer is 3.

Sample Input 2

2
1 2
1
1 2

Sample Output 2

1

The tree in this input is shown below:

All of the M restrictions will be satisfied only if Edge 1 is painted black, so the answer is 1.

Sample Input 3

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5

Sample Output 3

9

The tree in this input is shown below:

Sample Input 4

8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8

Sample Output 4

62

The tree in this input is shown below:

Solution in JavaScript (node.js v5.12) (judge verdict: Time Limit Exceeded)

"use strict";
var GET=(function(){function f(s){return new g(s);}function g(s){this._s=s.trim().split("\n");this._y=0;}g.prototype.a=function(f){var s=this._s, y=this._y, r;if(typeof s[y]==="string")s[y]=s[y].split(" ").reverse();r=s[y].pop();if(!s[y].length)this._y++;return f?r:+r;};g.prototype.l=function(f){var s=this._s[this._y++].split(" ");return f?s:s.map(a=>+a);};g.prototype.m=function(n,f){var r=this._s.slice(this._y,this._y+=n).map(a=>a.split(" "));return f?r:r.map(a=>a.map(a=>+a));};g.prototype.r=function(n,f){var r=this._s.slice(this._y,this._y+=n);return f?r:r.map(a=>+a);};return f;})();
var o=GET(require("fs").readFileSync("/dev/stdin","utf8"));
function xArray(v){var a=arguments,l=a.length,r="Array(a["+--l+"]).fill(0).map(x=>{return "+v+";})";while(--l)r="Array(a["+l+"]).fill(0).map(x=>"+r+")";return eval(r);}
var graph, bitor;

var myOut = main();
if(myOut !== undefined)console.log(myOut);
function main(){
  var n = o.a();
  var a = [], b = [];
  graph = xArray("[]",n);
  bitor = xArray(0,n-1);
  for(var i = 0; i < n-1; i++){
    var s = o.a()-1, t = o.a()-1;
    a[i] = Math.min(s,t), b[i] = Math.max(s,t);
    graph[s].push(t);
    graph[t].push(s);
  }
  var m = o.a();
  for(var i = 0; i < m; i++){
    var u = o.a()-1, v = o.a()-1;
    var bit = 1 << i;
    var vv = dfs(u,v,-1,[]);
    for(var j = 1; j < vv.length; j++){
      var p = Math.min(vv[j-1],vv[j]), q = vv[j-1]+vv[j]-p;
      for(var k = 0;;k++){
        if(a[k] === p && b[k] === q){
          bitor[k] |= bit;
          break;
        }
      }
    }
  }
  var mb = 1<<m;
  var dp = xArray(0,n,mb);
  dp[0][0] = 1;
  for(var i = 0; i < mb; i++){
    for(var j = 0; j < n-1; j++){
      if(bitor[j] === 0){
        dp[j+1][i] += dp[j][i]*2;
      }else{
        dp[j+1][i|bitor[j]] += dp[j][i];
        dp[j+1][i] += dp[j][i];
      }
    }
  }
  return dp[n-1][mb-1];
}

function dfs(s,g,p,V){
  V.push(s);
  if(g === s)return V;
  for(var v of graph[s]){
    if(v === p)continue;
    var q = dfs(v,g,s,V);
    if(q)return q;
  }
  V.pop();
}