Given a tree with n nodes numbered from 1 to n. Each node i has an associated value V_i.

If the simple path from u_1 to u_m consists of m nodes namely u_1 → u_2 → u_3 → ... u_{m-1} → u_{m}, then its alternating function A(u_{1},u_{m}) is defined as A(u_{1},u_{m}) = ∑_{i=1}^{m} (-1)^{i+1} ⋅ V_{u_{i}}. A path can also have 0 edges, i.e. u_{1}=u_{m}.

Compute the sum of alternating functions of all unique simple paths. Note that the paths are directed: two paths are considered different if the starting vertices differ or the ending vertices differ. The answer may be large so compute it modulo 10^{9}+7. 

Input

The first line contains an integer n (2 ≤ n ≤ 2⋅10^{5} ) — the number of vertices in the tree.

The second line contains n space-separated integers V_1, V_2, …, V_n (-10^9≤ V_i ≤ 10^9) — values of the nodes.

The next n-1 lines each contain two space-separated integers u and v (1≤ u, v≤ n, u ≠ v) denoting an edge between vertices u and v. It is guaranteed that the given graph is a tree.

Output

Print the total sum of alternating functions of all unique simple paths modulo 10^{9}+7. 

Examples

Input

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


Output

40


Input

8
-2 6 -4 -4 -9 -3 -7 23
8 2
2 3
1 4
6 5
7 6
4 7
5 8


Output

4

Note

Consider the first example.

A simple path from node 1 to node 2: 1 → 2 has alternating function equal to A(1,2) = 1 ⋅ (-4)+(-1) ⋅ 1 = -5.

A simple path from node 1 to node 3: 1 → 3 has alternating function equal to A(1,3) = 1 ⋅ (-4)+(-1) ⋅ 5 = -9.

A simple path from node 2 to node 4: 2 → 1 → 4 has alternating function A(2,4) = 1 ⋅ (1)+(-1) ⋅ (-4)+1 ⋅ (-2) = 3.

A simple path from node 1 to node 1 has a single node 1, so A(1,1) = 1 ⋅ (-4) = -4.

Similarly, A(2, 1) = 5, A(3, 1) = 9, A(4, 2) = 3, A(1, 4) = -2, A(4, 1) = 2, A(2, 2) = 1, A(3, 3) = 5, A(4, 4) = -2, A(3, 4) = 7, A(4, 3) = 7, A(2, 3) = 10, A(3, 2) = 10. So the answer is (-5) + (-9) + 3 + (-4) + 5 + 9 + 3 + (-2) + 2 + 1 + 5 + (-2) + 7 + 7 + 10 + 10 = 40.

Similarly A(1,4)=-2, A(2,2)=1, A(2,1)=5, A(2,3)=10, A(3,3)=5, A(3,1)=9, A(3,2)=10, A(3,4)=7, A(4,4)=-2, A(4,1)=2, A(4,2)=3 , A(4,3)=7 which sums upto 40. 
//make sure to make new file!
import java.io.*;
import java.util.*;

public class E960{
   
   public static int n;
   
   public static long MOD = 1000000007L;
   public static ArrayList<Integer>[] adj;
   
   public static int[] p;
   
   public static long[] oddc;
   public static long[] evenc;
   public static long[] oddp;
   public static long[] evenp;
   public static long[] add;
   public static long[] sub;
   
   public static void main(String[] args)throws IOException{
      BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
      PrintWriter out = new PrintWriter(System.out);
      
      n = Integer.parseInt(f.readLine());
      
      StringTokenizer st = new StringTokenizer(f.readLine());
      long[] array = new long[n+1];
      long sumarray = 0L;
      for(int k = 1; k <= n; k++){
         array[k] = Long.parseLong(st.nextToken());
         sumarray = (sumarray+array[k] + MOD)%MOD;
      }
      
      adj = new ArrayList[n+1];
      for(int k = 0; k <= n; k++) adj[k] = new ArrayList<Integer>();
      
      for(int k = 0; k < n-1; k++){
         st = new StringTokenizer(f.readLine());
         
         int a = Integer.parseInt(st.nextToken());
         int b = Integer.parseInt(st.nextToken());
         
         adj[a].add(b);
         adj[b].add(a);
      
      }
      
      p = new int[n+1];
      
      oddc = new long[n+1];
      evenc = new long[n+1];
      oddp = new long[n+1];
      evenp = new long[n+1];
      add = new long[n+1];
      sub = new long[n+1];
      
      dfsc1();
      dfsp1();
      dfsanswer1();
      
      
      /*
      dfsc(1,-1);
      dfsp(1,-1);
      dfsanswer(1,-1);
      */
      long answer = 0L;
      
      for(int k = 1; k <= n; k++){
         long prodadd = array[k]*(add[k]-sub[k]);
         prodadd = ((prodadd % MOD) + MOD) % MOD;
         answer = (answer + prodadd + MOD)%MOD;
      }
      
      answer = (answer*2 + MOD)%MOD;
      answer = (answer + sumarray + MOD)%MOD;
      out.println(answer);
         
      
      out.close();
   }
   
   public static void dfsc1(){
      
      
      Stack<Integer> stk = new Stack<Integer>();
      stk.add(-1);
      

      while(!stk.isEmpty()){
         int v = stk.pop();
         if(v < 0){
            stk.add(v*-1);
            v*=-1;
            for(int nei : adj[v]){
               if(nei == p[v]) 
                  continue;
               nei *= -1;
               
               stk.add(nei);
               p[nei*-1] = v;
            }
         } else {
            for(int nei : adj[v]){
               if(nei == p[v]) continue;
               oddc[v] += 1+evenc[nei];
               evenc[v] += oddc[nei];
            }
         }
      }   
      
   }
   
   
   public static void dfsp1(){
   
      Stack<Integer> stk = new Stack<Integer>();
      stk.add(1);
      
      
      while(!stk.isEmpty()){
         int v = stk.pop();
         if(p[v] > 0){
            oddp[v] = 1+evenp[p[v]]+evenc[p[v]]-oddc[v];
            evenp[v] = oddp[p[v]]+oddc[p[v]]-evenc[v]-1;
         }
      
         for(int nei : adj[v]){
            if(nei == p[v]) 
               continue;
            stk.add(nei);
            
         }
      }
   }
   
   public static void dfsanswer1(){
   
      
      Stack<Integer> stk = new Stack<Integer>();
      stk.add(1);
      
      while(!stk.isEmpty()){
         int v = stk.pop();
         add[v] = ((evenc[v]+1)*(evenp[v]+1)-1+MOD)%MOD;
         sub[v] = ((oddc[v])*(oddp[v])+MOD)%MOD;
         
         long sumoddc = 0L;
         long sumevenc = 0L;
         for(int nei : adj[v]){
            if(nei == p[v]) 
               continue;
            add[v] = (add[v] + sumoddc*oddc[nei]+MOD)%MOD;
            sub[v] = (sub[v] + sumevenc*(1+evenc[nei])+MOD)%MOD;
            sumoddc += oddc[nei];
            sumevenc += 1+evenc[nei];
            stk.add(nei);
         }
      }
   }
   
   /*
   //recursive versions, tle'd
   public static void dfsc(int v, int p){
      
      for(int nei : adj.get(v)){
         if(nei == p) 
            continue;
         dfsc(nei,v);
         oddc[v] += 1+evenc[nei];
         evenc[v] += oddc[nei];
      }
   }
   
   
   public static void dfsp(int v, int p){
      if(p != -1){
         oddp[v] = 1+evenp[p]+evenc[p]-oddc[v];
         evenp[v] = oddp[p]+oddc[p]-evenc[v]-1;
      }
      
      for(int nei : adj.get(v)){
         if(nei == p) 
            continue;
         dfsp(nei,v);
      }
   }
   
   public static void dfsanswer(int v, int p){
      add[v] = ((evenc[v]+1)*(evenp[v]+1)-1+MOD)%MOD;
      sub[v] = ((oddc[v])*(oddp[v])+MOD)%MOD;
      
      long sumoddc = 0L;
      long sumevenc = 0L;
      for(int nei : adj.get(v)){
         if(nei == p) 
            continue;
         add[v] = (add[v] + sumoddc*oddc[nei]+MOD)%MOD;
         sub[v] = (sub[v] + sumevenc*(1+evenc[nei])+MOD)%MOD;
         sumoddc += oddc[nei];
         sumevenc += 1+evenc[nei];
         dfsanswer(nei,v);
      }
   }
   */
      
}