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 = 201000;
const int MAXM = MAXN << 1;
bool vis[MAXN];
int a[MAXN], from[MAXN], to[MAXN], lv[MAXN], dfs_cnt;
int tree[MAXM];
vector<int> g[MAXN];
int lowbit(int x) { return x & (-x); }
void add(int pos, int val) {
  while (pos < MAXM) {
    tree[pos] += val;
    pos += lowbit(pos);
  }
}
int sum(int pos) {
  int ret = 0;
  while (pos > 0) {
    ret += tree[pos];
    pos -= lowbit(pos);
  }
  return ret;
}
void dfs(int u) {
  from[u] = ++dfs_cnt;
  vis[u] = true;
  int l = g[u].size();
  for (int i = 0; i < l; i++) {
    int v = g[u][i];
    if (!vis[v]) {
      lv[v] = lv[u] ^ 1;
      dfs(v);
    }
  }
  to[u] = ++dfs_cnt;
}
int main() {
  int n, m, i, t, u, v;
  while (scanf("%d%d", &n, &m) != EOF) {
    for (i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (i = 1; i < n; i++) {
      scanf("%d%d", &u, &v);
      g[u].push_back(v);
      g[v].push_back(u);
    }
    memset(vis, 0, sizeof(vis));
    dfs_cnt = 0;
    lv[1] = 0;
    dfs(1);
    memset(tree, 0, sizeof(tree));
    for (i = 1; i <= m; i++) {
      scanf("%d", &t);
      if (t == 1) {
        scanf("%d%d", &u, &v);
        if (lv[u]) {
          add(from[u], -v);
          add(to[u], v);
        } else {
          add(from[u], v);
          add(to[u], -v);
        }
      } else {
        scanf("%d", &u);
        if (lv[u])
          printf("%d\n", a[u] - sum(from[u]));
        else
          printf("%d\n", a[u] + sum(from[u]));
      }
    }
  }
  return 0;
}
