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;
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << '=' << h << '\n';
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << '=' << h << ',';
  _dbg(sdbg + 1, a...);
}
const int N = 2e5 + 7;
int n, m, a[N], par[N], in[N], out[N], id[N], timer;
vector<int> adj[N], g[2][N], nodes[2];
struct BIT {
  vector<int> bit;
  int n;
  BIT() {
    this->n = N + 5;
    bit.assign(this->n + 5, 0);
  }
  void add(int idx, int val) {
    for (++idx; idx < n; idx += idx & -idx) bit[idx] += val;
  }
  void range_add(int l, int r, int val) {
    add(l, val);
    add(r + 1, -val);
  }
  int query(int idx) {
    int ret = 0;
    for (++idx; idx > 0; idx -= idx & -idx) ret += bit[idx];
    return ret;
  }
} bt[2];
void init(int v, int lvl, int p) {
  if (p) par[v] = p;
  id[v] = lvl & 1;
  nodes[id[v]].push_back(v);
  for (auto ch : adj[v])
    if (ch != p) {
      g[id[p]][p].push_back(ch);
      init(ch, lvl + 1, v);
    }
}
void dfs(int v, int ii) {
  in[v] = ++timer;
  for (auto ch : g[ii][v]) dfs(ch, ii);
  out[v] = timer;
}
void add(int v, int val) { bt[id[v]].range_add(in[v], out[v], val); }
int query(int v) { return bt[id[v]].query(in[v]); }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  BIT bt[2];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  init(1, 0, 0);
  for (int i = 0; i < 2; i++) {
    timer = 0;
    for (auto v : nodes[i])
      if (!in[v]) dfs(v, i);
  }
  while (m--) {
    int q, x, val;
    cin >> q >> x;
    if (q == 1) {
      cin >> val;
      add(x, val);
    } else {
      cout << a[x] + query(x) - (par[x] ? query(par[x]) : 0) << '\n';
    }
  }
  return 0;
}
