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 = 412345;
int FWT[maxn];
int ar[maxn];
int vis[maxn];
int depth[maxn];
int in[maxn], out[maxn];
int cnt = 1;
int n, m, foo, bar, id, x, val;
vector<int> v[maxn];
void dfs(int n) {
  in[n] = cnt++;
  vis[n] = true;
  for (int i = 0; i < v[n].size(); i++) {
    int pt = v[n][i];
    if (!vis[pt]) {
      depth[pt] = depth[n] + 1;
      dfs(pt);
    }
  }
  out[n] = cnt++;
}
void update(int pos, int val) {
  while (pos < maxn) {
    FWT[pos] += val;
    pos += pos & (-pos);
  }
}
int query(int pos) {
  int r = 0;
  while (pos >= 1) {
    r += FWT[pos];
    pos -= pos & (-pos);
  }
  return r;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) scanf("%d", &ar[i + 1]);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &foo, &bar);
    v[foo].push_back(bar);
    v[bar].push_back(foo);
  }
  dfs(1);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &id, &x);
    if (id == 1) {
      scanf("%d", &val);
      if (depth[x] & 1) {
        update(in[x], -val);
        update(out[x], val);
      } else {
        update(in[x], val);
        update(out[x], -val);
      }
    } else {
      if (depth[x] & 1)
        printf("%d\n", ar[x] - query(in[x]));
      else
        printf("%d\n", ar[x] + query(in[x]));
    }
  }
  return 0;
}
