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 long double pi = acos(-1);
const int n_ = 2e5 + 5;
int n, q, st[n_ << 5], nodes[n_ << 2], f[n_], b[n_], arr[n_], cnt, odd[n_ << 5],
    even[n_ << 5];
vector<int> adj[n_];
void DFS(int u, int p = -1) {
  nodes[++cnt] = u;
  for (int i : adj[u]) {
    if (i != p) {
      DFS(i, u);
      nodes[++cnt] = u;
    }
  }
}
void upd(int parity, int s, int e, int v, int id = 1, int l = 1, int r = cnt) {
  if (s > r || e < l) return;
  if (s <= l && r <= e) {
    (parity ? odd : even)[id] += v;
    return;
  }
  upd(parity, s, e, v, id << 1, l, ((l + r) >> 1));
  upd(parity, s, e, v, id << 1 | 1, ((l + r) >> 1) + 1, r);
}
int get(int x, int id = 1, int l = 1, int r = cnt) {
  if (l == r) return arr[nodes[x]] + (odd[id] - even[id]) * (2 * (l & 1) - 1);
  if (odd[id]) {
    odd[id << 1] += odd[id];
    odd[id << 1 | 1] += odd[id];
  }
  if (even[id]) {
    even[id << 1] += even[id];
    even[id << 1 | 1] += even[id];
  }
  odd[id] = even[id] = 0;
  if (x <= ((l + r) >> 1))
    return get(x, id << 1, l, ((l + r) >> 1));
  else
    return get(x, id << 1 | 1, ((l + r) >> 1) + 1, r);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  DFS(1);
  for (int i = 1; i <= cnt; i++) b[nodes[i]] = i;
  for (int i = cnt; i; --i) f[nodes[i]] = i;
  int t, x, v;
  while (q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d %d", &x, &v);
      upd(f[x] % 2, f[x], b[x], v);
    } else {
      scanf("%d", &x);
      printf("%d\n", get(f[x]));
    }
  }
  return 0;
}
