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 N = 234567;
int ti;
int t[N + N];
int in[N + N];
int out[N + N];
int a[N];
int used[N];
int depth[N];
vector<int> g[N];
void dfs(int v) {
  used[v] = 1;
  in[v] = ti++;
  for (int i = 0; i < ((int)g[v].size()); i++) {
    int to = g[v][i];
    depth[to] = depth[v] + 1;
    if (!used[to]) dfs(to);
  }
  out[v] = ti++;
}
void upd(int pos, int d) {
  while (pos < 450000) t[pos] += d, pos |= pos + 1;
}
int get(int r) {
  int res = 0;
  while (r >= 0) res += t[r], r = (r & (r + 1)) - 1;
  return res;
}
int main() {
  int n, q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  for (int i = 0, f, t; i < n - 1; i++) {
    scanf("%d %d", &f, &t);
    --f, --t;
    g[f].push_back(t);
    g[t].push_back(f);
  }
  dfs(0);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      int pos, x;
      scanf("%d %d", &pos, &x);
      --pos;
      if (depth[pos] % 2 == 0)
        upd(in[pos], +x), upd(out[pos], -x);
      else
        upd(in[pos], -x), upd(out[pos], +x);
    } else {
      int pos;
      scanf("%d", &pos);
      --pos;
      int cur = get(in[pos]);
      if (depth[pos] % 2 == 0)
        printf("%d\n", a[pos] + cur);
      else
        printf("%d\n", a[pos] - cur);
    }
  }
  return 0;
}
