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>
struct node {
  int a;
  node* next;
};
int n, m, x, y, t, temp, v, a[200002], u[200002], b[200002], c[200002],
    order[200002], beg[200002], end[200002], level[200002], origins[200002],
    s = 0;
node* head[200002];
node* tail[200002];
void insertnode(int u, int v) {
  if (head[u] == NULL) {
    head[u] = new node;
    head[u]->a = v;
    head[u]->next = NULL;
    tail[u] = head[u];
  } else {
    node* p = new node;
    p->a = v;
    p->next = NULL;
    tail[u]->next = p;
    tail[u] = p;
  }
}
int lowbit(int i) { return i & (-i); }
void dfs(int i) {
  s++;
  u[i] = 1;
  order[s] = i;
  beg[i] = s;
  for (node* p = head[i]; p != NULL; p = p->next)
    if (u[p->a] == 0) {
      level[p->a] = level[i] + 1;
      dfs(p->a);
    }
  end[i] = s;
}
void iniz() {
  for (int i = 1; i <= n; i++) {
    c[i] = b[i];
    for (int j = i - 1; j > i - lowbit(i); j -= lowbit(j)) c[i] += c[j];
  }
  for (int i = 1; i <= n; i++) origins[i] = origins[i - 1] + b[i];
}
int sum(int k) {
  int s = 0;
  for (int i = k; i >= 1; i -= lowbit(i)) s += c[i];
  return s;
}
void modify(int pos, int v) {
  for (int i = pos; i <= n; i += lowbit(i)) c[i] += v;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    u[i] = 0;
    head[i] = NULL;
    tail[i] = NULL;
  }
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n - 1; i++) {
    scanf("%d%d", &x, &y);
    insertnode(x, y);
    insertnode(y, x);
  }
  level[1] = 1;
  dfs(1);
  for (int i = 1; i <= n; i++) b[i] = a[order[i]];
  iniz();
  for (int i = 1; i <= m; i++) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &x, &v);
      if (level[x] % 2 == 1) {
        modify(beg[x], v);
        modify(end[x] + 1, -v);
      } else {
        modify(beg[x], -v);
        modify(end[x] + 1, v);
      }
    } else {
      scanf("%d", &x);
      temp = sum(beg[x]) - origins[beg[x] - 1];
      if (level[x] % 2 == 1)
        printf("%d\n", temp);
      else
        printf("%d\n", a[x] * 2 - temp);
    }
  }
  return 0;
}
