#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 500, Mod = 1e9 + 7;
int gti(void) {
  char c = getchar();
  int ret = 0, st = 1;
  for (; !isdigit(c) && c != '-'; c = getchar())
    ;
  if (c == '-') st = -1, c = getchar();
  for (; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';
  return ret * st;
}
struct edge {
  edge *nxt;
  int to;
} pool[N * 2], *p = pool, *pt[N];
void add(int a, int b) {
  *(++p) = (edge){pt[a], b}, pt[a] = p;
  *(++p) = (edge){pt[b], a}, pt[b] = p;
}
int size[N], mx[N], done[N];
int getsize(int x) {
  size[x] = done[x] = true;
  for (edge *i = pt[x]; i; i = i->nxt)
    if (!done[i->to]) size[x] += getsize(i->to);
  done[x] = false;
  return size[x];
}
int getrt(int x, int sz) {
  int ret = 0;
  done[x] = true, mx[x] = sz - size[x];
  for (edge *i = pt[x]; i; i = i->nxt)
    if (!done[i->to]) {
      mx[x] = max(mx[x], size[i->to]);
      int pt = getrt(i->to, sz);
      if (!ret || mx[pt] < mx[ret]) ret = pt;
    }
  if (!ret || mx[x] < mx[ret]) ret = x;
  done[x] = false;
  return ret;
}
int c[2], f[2], cx[2], fx[2], v[N], n;
void count(int x, int xv, int t) {
  int now = (((long long)v[x] - xv) % Mod);
  f[t] = (((long long)f[t] + now) % Mod), c[t]++, done[x] = true;
  for (edge *i = pt[x]; i; i = i->nxt)
    if (!done[i->to]) count(i->to, now, t ^ 1);
  done[x] = false;
}
int getans(int x) {
  int ret = v[x];
  cx[0] = cx[1] = fx[0] = fx[1] = 0;
  for (edge *i = pt[x]; i; i = i->nxt)
    if (!done[i->to]) {
      c[0] = c[1] = f[0] = f[1] = 0;
      count(i->to, 0, 0);
      ret = (((long long)ret +
              (((long long)(((long long)(((long long)c[1] - c[0]) % Mod) *
                             (((long long)cx[1] + cx[0]) % Mod)) %
                            Mod) *
                v[x]) %
               Mod)) %
             Mod);
      ret = (((long long)ret +
              (((long long)(((long long)(((long long)cx[1] - cx[0]) % Mod) *
                             (((long long)c[1] + c[0]) % Mod)) %
                            Mod) *
                v[x]) %
               Mod)) %
             Mod);
      ret = (((long long)ret + (((long long)(((long long)cx[0] - cx[1]) % Mod) *
                                 (((long long)f[0] - f[1]) % Mod)) %
                                Mod)) %
             Mod);
      ret = (((long long)ret + (((long long)(((long long)c[0] + c[1]) % Mod) *
                                 (((long long)fx[0] + fx[1]) % Mod)) %
                                Mod)) %
             Mod);
      ret = (((long long)ret + (((long long)(((long long)c[0] - c[1]) % Mod) *
                                 (((long long)fx[0] - fx[1]) % Mod)) %
                                Mod)) %
             Mod);
      ret = (((long long)ret + (((long long)(((long long)cx[0] + cx[1]) % Mod) *
                                 (((long long)f[0] + f[1]) % Mod)) %
                                Mod)) %
             Mod);
      ret =
          (((long long)ret +
            (((long long)(((long long)(((long long)c[1] - c[0]) % Mod) * v[x]) %
                          Mod) +
              (((long long)f[1] + f[0]) % Mod)) %
             Mod)) %
           Mod);
      ret =
          (((long long)ret +
            (((long long)(((long long)(((long long)c[0] + c[1]) % Mod) * v[x]) %
                          Mod) +
              (((long long)f[1] - f[0]) % Mod)) %
             Mod)) %
           Mod);
      cx[0] = (((long long)cx[0] + c[0]) % Mod),
      fx[0] = (((long long)fx[0] + f[0]) % Mod);
      cx[1] = (((long long)cx[1] + c[1]) % Mod),
      fx[1] = (((long long)fx[1] + f[1]) % Mod);
    }
  return (((long long)ret + Mod) % Mod);
}
int solve(int rt, int sz) {
  rt = getrt(rt, sz), sz = getsize(rt);
  done[rt] = true;
  int ans = getans(rt);
  for (edge *i = pt[rt]; i; i = i->nxt)
    if (!done[i->to])
      ans = (((long long)ans + solve(i->to, size[i->to])) % Mod);
  return ans;
}
int main(void) {
  n = gti();
  for (int i = 1; i <= n; i++) v[i] = gti();
  for (int i = 1; i < n; i++) add(gti(), gti());
  printf("%d\n", solve(1, getsize(1)));
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 500, Mod = 1e9 + 7;
int gti(void) {
  char c = getchar();
  int ret = 0, st = 1;
  for (; !isdigit(c) && c != '-'; c = getchar())
    ;
  if (c == '-') st = -1, c = getchar();
  for (; isdigit(c); c = getchar()) ret = ret * 10 + c - '0';
  return ret * st;
}
struct edge {
  edge *nxt;
  int to;
} pool[N * 2], *p = pool, *pt[N];
void add(int a, int b) {
  *(++p) = (edge){pt[a], b}, pt[a] = p;
  *(++p) = (edge){pt[b], a}, pt[b] = p;
}
int size[N], mx[N], done[N];
int getsize(int x) {
  size[x] = done[x] = true;
  for (edge *i = pt[x]; i; i = i->nxt)
    if (!done[i->to]) size[x] += getsize(i->to);
  done[x] = false;
  return size[x];
}
int getrt(int x, int sz) {
  int ret = 0;
  done[x] = true, mx[x] = sz - size[x];
  for (edge *i = pt[x]; i; i = i->nxt)
    if (!done[i->to]) {
      mx[x] = max(mx[x], size[i->to]);
      int pt = getrt(i->to, sz);
      if (!ret || mx[pt] < mx[ret]) ret = pt;
    }
  if (!ret || mx[x] < mx[ret]) ret = x;
  done[x] = false;
  return ret;
}
int c[2], f[2], cx[2], fx[2], v[N], n;
void count(int x, int xv, int t) {
  int now = (((long long)v[x] - xv) % Mod);
  f[t] = (((long long)f[t] + now) % Mod), c[t]++, done[x] = true;
  for (edge *i = pt[x]; i; i = i->nxt)
    if (!done[i->to]) count(i->to, now, t ^ 1);
  done[x] = false;
}
int getans(int x) {
  int ret = v[x];
  cx[0] = cx[1] = fx[0] = fx[1] = 0;
  for (edge *i = pt[x]; i; i = i->nxt)
    if (!done[i->to]) {
      c[0] = c[1] = f[0] = f[1] = 0;
      count(i->to, 0, 0);
      ret = (((long long)ret +
              (((long long)(((long long)(((long long)c[1] - c[0]) % Mod) *
                             (((long long)cx[1] + cx[0]) % Mod)) %
                            Mod) *
                v[x]) %
               Mod)) %
             Mod);
      ret = (((long long)ret +
              (((long long)(((long long)(((long long)cx[1] - cx[0]) % Mod) *
                             (((long long)c[1] + c[0]) % Mod)) %
                            Mod) *
                v[x]) %
               Mod)) %
             Mod);
      ret = (((long long)ret + (((long long)(((long long)cx[0] - cx[1]) % Mod) *
                                 (((long long)f[0] - f[1]) % Mod)) %
                                Mod)) %
             Mod);
      ret = (((long long)ret + (((long long)(((long long)c[0] + c[1]) % Mod) *
                                 (((long long)fx[0] + fx[1]) % Mod)) %
                                Mod)) %
             Mod);
      ret = (((long long)ret + (((long long)(((long long)c[0] - c[1]) % Mod) *
                                 (((long long)fx[0] - fx[1]) % Mod)) %
                                Mod)) %
             Mod);
      ret = (((long long)ret + (((long long)(((long long)cx[0] + cx[1]) % Mod) *
                                 (((long long)f[0] + f[1]) % Mod)) %
                                Mod)) %
             Mod);
      ret =
          (((long long)ret +
            (((long long)(((long long)(((long long)c[1] - c[0]) % Mod) * v[x]) %
                          Mod) +
              (((long long)f[1] + f[0]) % Mod)) %
             Mod)) %
           Mod);
      ret =
          (((long long)ret +
            (((long long)(((long long)(((long long)c[0] + c[1]) % Mod) * v[x]) %
                          Mod) +
              (((long long)f[1] - f[0]) % Mod)) %
             Mod)) %
           Mod);
      cx[0] = (((long long)cx[0] + c[0]) % Mod),
      fx[0] = (((long long)fx[0] + f[0]) % Mod);
      cx[1] = (((long long)cx[1] + c[1]) % Mod),
      fx[1] = (((long long)fx[1] + f[1]) % Mod);
    }
  return (((long long)ret + Mod) % Mod);
}
int solve(int rt, int sz) {
  rt = getrt(rt, sz), sz = getsize(rt);
  done[rt] = true;
  int ans = getans(rt);
  for (edge *i = pt[rt]; i; i = i->nxt)
    if (!done[i->to])
      ans = (((long long)ans + solve(i->to, size[i->to])) % Mod);
  return ans;
}
int main(void) {
  n = gti();
  for (int i = 1; i <= n; i++) v[i] = gti();
  for (int i = 1; i < n; i++) add(gti(), gti());
  printf("%d\n", solve(1, getsize(1)));
  return 0;
}

```