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 long long int N = 1000010;
long long int a1[N];
vector<long long int> v1[N];
struct lol {
  long long int in, out, on;
};
lol G[N];
long long int t = 1, vc[2][N];
long long int n, q;
void dfs(long long int u, long long int par) {
  G[u].in = t++;
  G[u].on = (par == -1) ? 0 : (1 - G[par].on);
  vector<long long int>::iterator it;
  for (it = v1[u].begin(); it != v1[u].end(); it++) {
    if (*it == par) continue;
    dfs(*it, u);
  }
  G[u].out = t++;
}
void updt(long long int pos, long long int on, long long int val) {
  while (pos <= 2 * n) {
    vc[on][pos] += val;
    pos += (pos & -pos);
  }
}
long long int qry(long long int pos, long long int on) {
  long long int res = 0;
  while (pos > 0) {
    res += vc[on][pos];
    pos -= (pos & -pos);
  }
  return res;
}
int main() {
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> a1[i];
  for (int i = 1; i < n; i++) {
    long long int u, v;
    cin >> u >> v;
    v1[u].push_back(v);
    v1[v].push_back(u);
  }
  dfs(1, -1);
  for (int i = 1; i <= n; i++) {
  }
  while (q--) {
    long long int type;
    cin >> type;
    long long int u;
    cin >> u;
    if (type == 1) {
      long long int val;
      cin >> val;
      updt(G[u].in, G[u].on, val);
      updt(G[u].out, G[u].on, -val);
    } else {
      cout << a1[u] + (qry(G[u].in, G[u].on)) - qry(G[u].in, 1 - G[u].on)
           << "\n";
    }
  }
  return 0;
}
