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 N = 2e5 + 10;
vector<int> g[N];
int a[N];
int n, q;
int f[N];
void add(int id, int val) {
  for (; id < N; id += (id & -id)) f[id] += val;
}
int sum(int x) {
  int res = 0;
  for (; x > 0; x -= (x & -x)) res += f[x];
  return res;
}
int tin[N];
int tout[N];
bool h[N];
int timer = 1;
void dfs(int node, int par = -1, int dep = 0) {
  h[node] = dep;
  tin[node] = tout[node] = timer++;
  for (auto x : g[node])
    if (x != par) dfs(x, node, dep ^ 1), tout[node] = tout[x];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> q;
  for (int i = 0; i < n; i++) cin >> a[i + 1];
  int u, v, x;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  for (int i = 0; i < q; i++) {
    cin >> x >> u;
    if (x == 1) {
      cin >> v;
      if (h[u]) v *= -1;
      add(tin[u], v);
      add(tout[u] + 1, -v);
    } else {
      int res = sum(tin[u]);
      if (h[u]) res = -res;
      res += a[u];
      cout << res << '\n';
    }
  }
}
