#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
using namespace std;
const long long mo = 1e9 + 7;
template <typename _T>
void read(_T &x) {
  _T f = 1;
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
long long sum1[200005], sum2[200005], ans, val[200005];
long long siz[200005], sup1[200005], sup2[200005];
int head[200005], tot, dep[200005], n, id[200005], f[200005];
struct edge {
  int v, nxt;
} e[200005 << 1];
bool cmp(int a, int b) { return dep[a] < dep[b]; }
void addEdge(int u, int v) {
  e[++tot] = (edge){v, head[u]};
  head[u] = tot;
}
void dfs(int u, int fa) {
  f[u] = fa;
  dep[u] = dep[fa] + 1;
  sum1[u]++;
  siz[u]++;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    dfs(v, u);
    sum1[u] += sum2[v];
    sum2[u] += sum1[v];
    siz[u] += siz[v];
  }
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    (ans +=
     (sum2[v] - sum1[v]) % mo * (siz[u] - siz[v] - 1) % mo * val[u] % mo) %= mo;
  }
  (ans += (siz[u] - 1) * val[u] % mo) %= mo;
  (ans += (sum1[u] - sum2[u]) % mo * (n - siz[u] + 1) % mo * val[u] % mo) %= mo;
}
signed main() {
  read(n);
  for (int i = 1; i <= n; i++)
    read(val[i]), val[i] = (val[i] + mo) % mo, id[i] = i;
  for (int i = 1; i < n; i++) {
    int u, v;
    read(u);
    read(v);
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs(1, 0);
  sort(id + 1, id + n + 1, cmp);
  for (int i = 2; i <= n; i++) {
    int u = id[i];
    sup1[u] = sup2[f[u]] + sum2[f[u]] - sum1[u];
    sup2[u] = sup1[f[u]] + sum1[f[u]] - sum2[u];
    (ans += (sup1[u] - sup2[u]) % mo * siz[u] % mo * val[u] % mo) %= mo;
  }
  printf("%lld\n", (ans + mo) % mo);
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
using namespace std;
const long long mo = 1e9 + 7;
template <typename _T>
void read(_T &x) {
  _T f = 1;
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9') {
    x = (x << 3) + (x << 1) + (s ^ 48);
    s = getchar();
  }
  x *= f;
}
long long sum1[200005], sum2[200005], ans, val[200005];
long long siz[200005], sup1[200005], sup2[200005];
int head[200005], tot, dep[200005], n, id[200005], f[200005];
struct edge {
  int v, nxt;
} e[200005 << 1];
bool cmp(int a, int b) { return dep[a] < dep[b]; }
void addEdge(int u, int v) {
  e[++tot] = (edge){v, head[u]};
  head[u] = tot;
}
void dfs(int u, int fa) {
  f[u] = fa;
  dep[u] = dep[fa] + 1;
  sum1[u]++;
  siz[u]++;
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    dfs(v, u);
    sum1[u] += sum2[v];
    sum2[u] += sum1[v];
    siz[u] += siz[v];
  }
  for (int i = head[u]; i; i = e[i].nxt) {
    int v = e[i].v;
    if (v == fa) continue;
    (ans +=
     (sum2[v] - sum1[v]) % mo * (siz[u] - siz[v] - 1) % mo * val[u] % mo) %= mo;
  }
  (ans += (siz[u] - 1) * val[u] % mo) %= mo;
  (ans += (sum1[u] - sum2[u]) % mo * (n - siz[u] + 1) % mo * val[u] % mo) %= mo;
}
signed main() {
  read(n);
  for (int i = 1; i <= n; i++)
    read(val[i]), val[i] = (val[i] + mo) % mo, id[i] = i;
  for (int i = 1; i < n; i++) {
    int u, v;
    read(u);
    read(v);
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs(1, 0);
  sort(id + 1, id + n + 1, cmp);
  for (int i = 2; i <= n; i++) {
    int u = id[i];
    sup1[u] = sup2[f[u]] + sum2[f[u]] - sum1[u];
    sup2[u] = sup1[f[u]] + sum1[f[u]] - sum2[u];
    (ans += (sup1[u] - sup2[u]) % mo * siz[u] % mo * val[u] % mo) %= mo;
  }
  printf("%lld\n", (ans + mo) % mo);
  return 0;
}

```