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 M = 2e5 + 5;
int n, q;
int seg[4 * M], lazy[4 * M][2], t[M];
vector<int> adj[M];
int st[M], en[M], id[M], hi[M];
int dfsnum;
int s, e, val, p;
void dfs(int x, int p) {
  st[x] = dfsnum;
  id[dfsnum++] = x;
  hi[x] = hi[p] + 1;
  for (int i : adj[x]) {
    if (i == p) continue;
    dfs(i, x);
  }
  en[x] = dfsnum - 1;
}
void fix(int nd, int l, int r) {
  if (l != r) {
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++) lazy[nd * 2 + i][j] += lazy[nd][j];
  } else {
    seg[nd] += lazy[nd][hi[id[l]] % 2];
    seg[nd] -= lazy[nd][1 - hi[id[l]] % 2];
  }
  lazy[nd][0] = lazy[nd][1] = 0;
}
void build(int nd, int l, int r) {
  if (l == r) {
    seg[nd] += t[id[l]];
    return;
  }
  int mid = (l + r) / 2;
  build(nd * 2, l, mid);
  build(nd * 2 + 1, mid + 1, r);
}
void update(int nd, int l, int r) {
  fix(nd, l, r);
  if (r < s || e < l) return;
  if (s <= l && r <= e) {
    lazy[nd][p] += val;
    fix(nd, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(nd * 2, l, mid);
  update(nd * 2 + 1, mid + 1, r);
}
int get(int nd, int l, int r) {
  fix(nd, l, r);
  if (r < s || e < l) return 0;
  if (s <= l && r <= e) return seg[nd];
  int mid = (l + r) / 2;
  return get(nd * 2, l, mid) + get(nd * 2 + 1, mid + 1, r);
}
int main() {
  scanf("%d %d", &n, &q);
  for (int i = 1; i <= n; i++) scanf("%d", &t[i]);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  build(1, 0, n - 1);
  while (q--) {
    int t, x;
    scanf("%d %d", &t, &x);
    if (t == 1) {
      int v;
      scanf("%d", &v);
      s = st[x], e = en[x], val = v, p = hi[x] % 2;
      update(1, 0, n - 1);
    } else {
      s = st[x], e = st[x];
      printf("%d\n", get(1, 0, n - 1));
    }
  }
}
