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 lt = 200005;
int strength[lt];
vector<int> adj[lt];
bool visited[lt];
vector<int> arr;
vector<int> st(lt);
vector<int> en(lt);
vector<int> level(lt);
vector<int> segodd(4 * lt);
vector<int> segeven(4 * lt);
int tp = 0;
int sum = 0;
void dfs(int u, int l) {
  arr.push_back(u);
  st[u] = tp;
  level[u] = l;
  tp++;
  visited[u] = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (!visited[v]) dfs(v, l + 1);
  }
  en[u] = (tp - 1);
}
void add(int s, int e, int l, int r, int val, int isOdd, int i) {
  if (e < l || s > r) return;
  if (l <= s && e <= r) {
    if (isOdd)
      segodd[i] += val;
    else
      segeven[i] += val;
    return;
  } else {
    int m = (s + e) / 2;
    add(s, m, l, r, val, isOdd, 2 * i + 1);
    add(m + 1, e, l, r, val, isOdd, 2 * i + 2);
  }
}
void query(int s, int e, int x, int isOdd, int i) {
  if (x < s || x > e) return;
  if (s <= x && x <= e) {
    if (isOdd)
      sum += segodd[i] - segeven[i];
    else
      sum += segeven[i] - segodd[i];
  }
  if (s == e) return;
  int m = (s + e) / 2;
  query(s, m, x, isOdd, 2 * i + 1);
  query(m + 1, e, x, isOdd, 2 * i + 2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> strength[i];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  while (m--) {
    int q, x, v;
    cin >> q;
    if (q == 1) {
      cin >> x >> v;
      add(0, n - 1, st[x], en[x], v, level[x] % 2, 0);
    } else if (q == 2) {
      cin >> x;
      sum = 0;
      query(0, n - 1, st[x], level[x] % 2, 0);
      cout << strength[x - 1] + sum << "\n";
    }
  }
  return 0;
}
