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 = 200005;
vector<vector<int>> g(200005);
int n, m;
long long int tme = 1, start[N], End[N], bit[2 * N], arr[N], level[N];
void update(int idx, long long int inc) {
  while (idx <= 2 * N) {
    bit[idx] += inc;
    idx += (idx & -idx);
  }
}
long long int query(int idx) {
  long long int ans = 0;
  while (idx > 0) {
    ans += bit[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
void dfs(int cur, int par, int d) {
  start[cur] = tme++;
  level[cur] = d;
  for (auto v : g[cur]) {
    if (v == par) continue;
    dfs(v, cur, d + 1);
  }
  End[cur] = tme++;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0, 0);
  while (m--) {
    long long int num, x, val;
    cin >> num;
    if (num == 1) {
      cin >> x >> val;
      if (level[x] % 2) val = -val;
      update(start[x], val);
      update(End[x], -val);
    } else {
      cin >> x;
      long long int ans;
      if (level[x] % 2 == 0)
        ans = arr[x] + query(start[x]);
      else
        ans = arr[x] - query(start[x]);
      cout << ans << "\n";
    }
  }
  return 0;
}
