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 head[200000 * 3], top = 0, num = 1;
struct node {
  int to;
  int next;
} edge[200000 * 3];
int a[200000 * 3];
long long bit[3][200000 * 3];
long long s[4][200000 * 3];
void add(int from, int to) {
  edge[top].to = to;
  edge[top].next = head[from];
  head[from] = top++;
}
void dfs(int now, int f, int z) {
  s[0][now] = z;
  s[1][now] = ++num;
  for (int k = head[now]; k != -1; k = edge[k].next)
    if (edge[k].to != f) dfs(edge[k].to, now, 1 - z);
  s[2][now] = num;
}
void add_1(int x, int i, int v) {
  while (i <= 4000003) {
    bit[x][i] += v;
    i += i & -i;
  }
}
long long sum(int x, int i) {
  long long ans = 0;
  while (i > 0) {
    ans += bit[x][i];
    i -= i & -i;
  }
  return ans;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n - 1; i++) {
    int ss, b;
    scanf("%d%d", &ss, &b);
    add(ss, b);
    add(b, ss);
  }
  dfs(1, -1, 0);
  for (int i = 1; i <= m; i++) {
    int x;
    int u, va;
    scanf("%d", &x);
    if (x == 1) {
      scanf("%d%d", &u, &va);
      add_1(s[0][u], s[1][u], va);
      add_1(s[0][u], s[2][u] + 1, -va);
      add_1((s[0][u] + 1) % 2, s[1][u], -va);
      add_1((s[0][u] + 1) % 2, s[2][u] + 1, va);
    } else {
      scanf("%d", &u);
      printf("%I64d\n", a[u] + sum(s[0][u], s[1][u]));
    }
  }
  return 0;
}
