Iahub likes trees very much. Recently he discovered an interesting tree named propagating tree. The tree consists of n nodes numbered from 1 to n, each node i having an initial value ai. The root of the tree is node 1.

This tree has a special property: when a value val is added to a value of node i, the value -val is added to values of all the children of node i. Note that when you add value -val to a child of node i, you also add -(-val) to all children of the child of node i and so on. Look an example explanation to understand better how it works.

This tree supports two types of queries:

  * "1 x val" — val is added to the value of node x; 
  * "2 x" — print the current value of node x. 



In order to help Iahub understand the tree better, you must answer m queries of the preceding type.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 200000). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 1000). Each of the next n–1 lines contains two integers vi and ui (1 ≤ vi, ui ≤ n), meaning that there is an edge between nodes vi and ui.

Each of the next m lines contains a query in the format described above. It is guaranteed that the following constraints hold for all queries: 1 ≤ x ≤ n, 1 ≤ val ≤ 1000.

Output

For each query of type two (print the value of node x) you must print the answer to the query on a separate line. The queries must be answered in the order given in the input.

Examples

Input

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


Output

3
3
0

Note

The values of the nodes are [1, 2, 1, 1, 2] at the beginning.

Then value 3 is added to node 2. It propagates and value -3 is added to it's sons, node 4 and node 5. Then it cannot propagate any more. So the values of the nodes are [1, 5, 1, - 2, - 1].

Then value 2 is added to node 1. It propagates and value -2 is added to it's sons, node 2 and node 3. From node 2 it propagates again, adding value 2 to it's sons, node 4 and node 5. Node 3 has no sons, so it cannot propagate from there. The values of the nodes are [3, 3, - 1, 0, 1].

You can see all the definitions about the tree at the following link: http://en.wikipedia.org/wiki/Tree_(graph_theory)
#include <bits/stdc++.h>
using namespace std;
const int c = 2e5 + 5;
int strength[c], level[c], size[c], visited[c], tree_even[4 * c],
    tree_odd[4 * c], M;
std::vector<int> v[c], preorder;
map<int, int> Map;
int dfs(int vertex, int lev) {
  if (visited[vertex]) return 0;
  visited[vertex] = 1;
  level[vertex] = lev;
  size[vertex] = 1;
  preorder.push_back(vertex);
  Map[vertex] = (int)preorder.size() - 1;
  for (int i = 0; i < (int)v[vertex].size(); i++) {
    size[vertex] += dfs(v[vertex][i], lev + 1);
  }
  return size[vertex];
}
void update(int index, int s, int e, int l, int r, int val) {
  if (e < l || s > r) return;
  if (s >= l && e <= r) {
    if (M == 0)
      tree_even[index] += val;
    else
      tree_odd[index] += val;
    return;
  }
  int mid = (s + e) / 2;
  update(2 * index, s, mid, l, r, val);
  update(2 * index + 1, mid + 1, e, l, r, val);
  return;
}
int sum;
int query(int index, int l, int r, int i) {
  if (i < l || i > r) return 0;
  if (l <= i && i <= r) {
    if (M == 0)
      sum += tree_even[index] - tree_odd[index];
    else
      sum += tree_odd[index] - tree_even[index];
  }
  if (l == r) return 0;
  int mid = (l + r) / 2;
  query(2 * index, l, mid, i);
  query(2 * index + 1, mid + 1, r, i);
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  int n, m, v1, v2, type, node, upd, start;
  cin >> n >> m;
  for (int i = 1; i < n + 1; i++) cin >> strength[i];
  for (int i = 0; i < n - 1; i++) {
    cin >> v1 >> v2;
    v[v1].push_back(v2);
    v[v2].push_back(v1);
  }
  preorder.push_back(0);
  dfs(1, 0);
  while (m--) {
    cin >> type;
    if (type == 1) {
      cin >> node >> upd;
      start = Map[node];
      M = level[node] % 2;
      update(1, 1, n, start, start + size[node] - 1, upd);
    } else {
      cin >> node;
      start = Map[node];
      M = level[node] % 2;
      sum = strength[node];
      query(1, 1, n, start);
      cout << sum << "\n";
    }
  }
  return 0;
}
