#include <bits/stdc++.h>
using namespace std;
const long long mxN = 200005;
vector<long long> adj[mxN];
long long a[mxN];
long long sz[mxN];
long long cpar[mxN];
bool vis[mxN];
const long long MOD = 1e9 + 7;
long long add(long long a, long long b) {
  a += b;
  if (a < 0) {
    a += MOD;
  }
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
long long mult(long long a, long long b) { return ((long long)a * b) % MOD; }
void dfs(long long n, long long p = -1) {
  sz[n] = 1;
  for (long long v : adj[n])
    if (v != p && !vis[v]) {
      dfs(v, n);
      sz[n] += sz[v];
    }
}
long long centroid(long long n) {
  dfs(n);
  long long num = sz[n];
  long long p = -1;
  do {
    long long nxt = -1;
    for (long long v : adj[n])
      if (v != p && !vis[v]) {
        if (2 * sz[v] > num) nxt = v;
      }
    p = n, n = nxt;
  } while (~n);
  return p;
}
long long val[mxN][3];
long long depth[mxN];
long long ans;
void dep(long long v, long long p) {
  val[v][0] = val[v][1] = val[v][2] = 0;
  for (long long i : adj[v]) {
    if (i == p || vis[i]) continue;
    depth[i] = depth[v] + 1;
    dep(i, v);
  }
}
void dfs1(long long v, long long p, long long pos, long long neg,
          long long alt) {
  alt = add(mult(alt, -1), a[v]);
  if (depth[v] % 2) {
    pos = sub(pos, a[v]);
    neg = add(neg, a[v]);
  } else {
    pos = add(pos, a[v]);
    neg = sub(neg, a[v]);
    ans = add(ans, mult(2, pos));
  }
  val[v][0] = pos;
  val[v][1] = neg;
  val[v][2] = alt;
  for (long long i : adj[v]) {
    if (i == p || vis[i]) continue;
    dfs1(i, v, pos, neg, alt);
  }
}
long long cur[2];
long long tot;
void update(long long v, long long p) {
  ++tot;
  cur[0] = add(cur[0], val[v][0]);
  cur[1] = add(cur[1], val[v][1]);
  for (long long i : adj[v]) {
    if (i == p || vis[i]) continue;
    update(i, v);
  }
}
void solve(long long v, long long p) {
  if (depth[v] % 2) {
    ans = add(ans, add(cur[1], mult(tot, val[v][2])));
  } else {
    ans = add(ans, add(cur[0], mult(tot, val[v][2])));
  }
  for (long long i : adj[v]) {
    if (i == p || vis[i]) continue;
    solve(i, v);
  }
}
void centroid_decomp(long long n = 0, long long p = -1) {
  long long c = centroid(n);
  ans = add(ans, a[c]);
  depth[c] = 0;
  dep(c, -1);
  vector<long long> cs;
  for (long long i : adj[c]) {
    if (vis[i]) continue;
    cs.push_back(i);
    dfs1(i, c, a[c], -a[c], 0);
  }
  tot = cur[0] = cur[1] = 0;
  for (long long i : cs) {
    solve(i, c);
    update(i, c);
  }
  reverse((cs).begin(), (cs).end());
  tot = cur[0] = cur[1] = 0;
  for (long long i : cs) {
    solve(i, c);
    update(i, c);
  }
  vis[c] = true;
  cpar[c] = p;
  for (long long v : adj[c])
    if (!vis[v]) {
      centroid_decomp(v, c);
    }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < n - 1; ++i) {
    long long a, b;
    cin >> a >> b;
    --a;
    --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  centroid_decomp();
  cout << ans << "\n";
}

### Prompt

Create a solution in CPP for 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>
using namespace std;
const long long mxN = 200005;
vector<long long> adj[mxN];
long long a[mxN];
long long sz[mxN];
long long cpar[mxN];
bool vis[mxN];
const long long MOD = 1e9 + 7;
long long add(long long a, long long b) {
  a += b;
  if (a < 0) {
    a += MOD;
  }
  if (a >= MOD) {
    a -= MOD;
  }
  return a;
}
long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += MOD;
  return a;
}
long long mult(long long a, long long b) { return ((long long)a * b) % MOD; }
void dfs(long long n, long long p = -1) {
  sz[n] = 1;
  for (long long v : adj[n])
    if (v != p && !vis[v]) {
      dfs(v, n);
      sz[n] += sz[v];
    }
}
long long centroid(long long n) {
  dfs(n);
  long long num = sz[n];
  long long p = -1;
  do {
    long long nxt = -1;
    for (long long v : adj[n])
      if (v != p && !vis[v]) {
        if (2 * sz[v] > num) nxt = v;
      }
    p = n, n = nxt;
  } while (~n);
  return p;
}
long long val[mxN][3];
long long depth[mxN];
long long ans;
void dep(long long v, long long p) {
  val[v][0] = val[v][1] = val[v][2] = 0;
  for (long long i : adj[v]) {
    if (i == p || vis[i]) continue;
    depth[i] = depth[v] + 1;
    dep(i, v);
  }
}
void dfs1(long long v, long long p, long long pos, long long neg,
          long long alt) {
  alt = add(mult(alt, -1), a[v]);
  if (depth[v] % 2) {
    pos = sub(pos, a[v]);
    neg = add(neg, a[v]);
  } else {
    pos = add(pos, a[v]);
    neg = sub(neg, a[v]);
    ans = add(ans, mult(2, pos));
  }
  val[v][0] = pos;
  val[v][1] = neg;
  val[v][2] = alt;
  for (long long i : adj[v]) {
    if (i == p || vis[i]) continue;
    dfs1(i, v, pos, neg, alt);
  }
}
long long cur[2];
long long tot;
void update(long long v, long long p) {
  ++tot;
  cur[0] = add(cur[0], val[v][0]);
  cur[1] = add(cur[1], val[v][1]);
  for (long long i : adj[v]) {
    if (i == p || vis[i]) continue;
    update(i, v);
  }
}
void solve(long long v, long long p) {
  if (depth[v] % 2) {
    ans = add(ans, add(cur[1], mult(tot, val[v][2])));
  } else {
    ans = add(ans, add(cur[0], mult(tot, val[v][2])));
  }
  for (long long i : adj[v]) {
    if (i == p || vis[i]) continue;
    solve(i, v);
  }
}
void centroid_decomp(long long n = 0, long long p = -1) {
  long long c = centroid(n);
  ans = add(ans, a[c]);
  depth[c] = 0;
  dep(c, -1);
  vector<long long> cs;
  for (long long i : adj[c]) {
    if (vis[i]) continue;
    cs.push_back(i);
    dfs1(i, c, a[c], -a[c], 0);
  }
  tot = cur[0] = cur[1] = 0;
  for (long long i : cs) {
    solve(i, c);
    update(i, c);
  }
  reverse((cs).begin(), (cs).end());
  tot = cur[0] = cur[1] = 0;
  for (long long i : cs) {
    solve(i, c);
    update(i, c);
  }
  vis[c] = true;
  cpar[c] = p;
  for (long long v : adj[c])
    if (!vis[v]) {
      centroid_decomp(v, c);
    }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n;
  cin >> n;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < n - 1; ++i) {
    long long a, b;
    cin >> a >> b;
    --a;
    --b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  centroid_decomp();
  cout << ans << "\n";
}

```