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 maxn = 5e5 + 10;
vector<int> adj[maxn];
int n, a[maxn], q;
void input() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v), adj[v].push_back(u);
  }
}
int T = 0;
bool mark[maxn];
int st[maxn], en[maxn], h[maxn];
void dfs(int v = 0) {
  mark[v] = 1;
  st[v] = T;
  for (int u : adj[v])
    if (!mark[u]) {
      h[u] = h[v] + 1;
      T++;
      dfs(u);
    }
  en[v] = T;
}
int what_the_hell_I_Dont_have_any_idea_about_this_variable[4 * maxn];
void add(int x, int val, int lq, int rq, int node = 0, int lnode = 0,
         int rnode = n) {
  if (rq <= lnode || rnode <= lq) return;
  if (lq <= lnode && rnode <= rq) {
    what_the_hell_I_Dont_have_any_idea_about_this_variable[node] += val;
    return;
  }
  int md = (lnode + rnode) / 2;
  add(x, val, lq, rq, 2 * node + 1, lnode, md);
  add(x, val, lq, rq, 2 * node + 2, md, rnode);
}
int get(int x, int node = 0, int lnode = 0, int rnode = n) {
  if (st[x] < lnode || rnode <= st[x]) return 0;
  if (rnode - lnode == 1)
    return what_the_hell_I_Dont_have_any_idea_about_this_variable[node];
  int res = 0, md = (lnode + rnode) / 2;
  res += get(x, node * 2 + 1, lnode, md);
  res += get(x, node * 2 + 2, md, rnode);
  return res + what_the_hell_I_Dont_have_any_idea_about_this_variable[node];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cin.tie(0);
  input();
  dfs();
  for (; q--;) {
    int t;
    cin >> t;
    if (t == 1) {
      int x, val;
      cin >> x >> val;
      x--;
      if (h[x] % 2) val *= -1;
      add(x, val, st[x], en[x] + 1);
    } else {
      int x;
      cin >> x;
      x--;
      int ans = get(x);
      if (h[x] % 2) ans *= -1;
      printf("%d\n", ans + a[x]);
    }
  }
}
