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;
int a[200001];
int n, m;
int cmd, x, val;
int fenwick[200001];
int rt[200001];
bool d[200001];
int pos[200001];
vector<int> g[200001];
int timer;
void update(int i, int val) {
  for (; i < n; i |= i + 1) fenwick[i] += val;
}
int sum(int r) {
  int res;
  for (res = 0; r > -1; r = (r & (r + 1)) - 1) res += fenwick[r];
  return res;
}
void dfs(int v, bool f = false, int p = -1) {
  d[v] = f;
  pos[v] = timer;
  update(timer, f ? -a[v] : a[v]);
  update(++timer, f ? a[v] : -a[v]);
  for (int i = 0; i < g[v].size(); ++i)
    if (g[v][i] != p) dfs(g[v][i], !f, v);
  rt[v] = timer;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  for (int i = 1, u, v; i < n; ++i) {
    scanf("%d %d", &u, &v);
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  while (m--) {
    scanf("%d %d", &cmd, &x);
    --x;
    if (cmd == 1) {
      scanf("%d", &val);
      if (d[x]) val = -val;
      update(pos[x], val);
      update(rt[x], -val);
    } else
      printf("%d\n", d[x] ? -sum(pos[x]) : sum(pos[x]));
  }
  return 0;
}
