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 const N = 2e5 + 10;
int n, m, a[N], b[N], s_time[N], f_time[N], t, seg[N << 2][2], ans;
bool dis[N];
vector<int> adj[N];
void read_input() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
}
void dfs(int v) {
  s_time[v] = ++t;
  for (int u : adj[v])
    if (!s_time[u]) {
      dis[u] = !dis[v];
      dfs(u);
    }
  f_time[v] = t;
}
void add(int L, int R, bool d, int val, int st = 1, int en = n + 1,
         int id = 1) {
  if (en <= L || st >= R) return;
  if (L <= st && en <= R) {
    seg[id][d] += val, seg[id][!d] -= val;
    return;
  }
  int mid = (st + en) >> 1;
  add(L, R, d, val, st, mid, id << 1), add(L, R, d, val, mid, en, id << 1 | 1);
}
void get(int p, bool d, int st = 1, int en = n + 1, int id = 1) {
  if (st > p || en <= p) return;
  ans += seg[id][d];
  if (en - st == 1) return;
  int mid = (st + en) >> 1;
  if (p < mid)
    get(p, d, st, mid, id << 1);
  else
    get(p, d, mid, en, id << 1 | 1);
}
void write_output() {
  while (m--) {
    int opr, x, val;
    cin >> opr >> x;
    x--;
    if (opr == 1) {
      cin >> val;
      add(s_time[x], f_time[x] + 1, dis[x], val);
    } else {
      ans = 0;
      get(s_time[x], dis[x]);
      cout << ans + a[x] << '\n';
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  read_input(), dfs(0), write_output();
  return 0;
}
