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;
long long n, m, x, y, typ, val, timer, ps = 1ll, disc[201000], tin[201000],
                                       tout[201000], a[210000], ev[210000],
                                       od[210000];
vector<long long> v[201000];
bool used[210000], even[210000];
void dfs(long long, long long);
void update(long long, long long);
long long sum(long long);
int main() {
  cin >> n >> m;
  for (long long i = 1ll; i <= n; i++) cin >> a[i];
  for (long long i = 1ll; i < n; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1ll, 1ll);
  for (long long i = 1ll; i <= m; i++) {
    cin >> typ;
    if (typ == 1ll) {
      cin >> x >> val;
      if (even[x])
        update(x, val);
      else
        update(x, -val);
    } else {
      cin >> x;
      cout << sum(x) << endl;
    }
  }
  cin >> n;
  return 0;
}
void dfs(long long pos, long long lvl) {
  if (lvl % 2ll == 0ll) even[pos] = true;
  used[pos] = true;
  disc[ps] = pos;
  ps++;
  timer++;
  tin[pos] = timer;
  for (long long i = 0ll; i < v[pos].size(); i++)
    if (!used[v[pos][i]]) dfs(v[pos][i], lvl + 1ll);
  tout[pos] = timer;
  return;
}
void update(long long pos, long long vl) {
  long long l = tin[pos], r = tout[pos] + 1ll;
  for (; l <= n; l |= (l + 1ll)) {
    ev[l] += vl;
    od[l] -= vl;
  }
  for (; r <= n; r |= (r + 1ll)) {
    ev[r] -= vl;
    od[r] += vl;
  }
}
long long sum(long long pos) {
  long long ret = a[pos], i = tin[pos];
  bool f;
  if (even[pos])
    f = true;
  else
    f = false;
  for (; i >= 1; i = (i & (i + 1ll)) - 1ll) {
    if (f)
      ret += ev[i];
    else
      ret += od[i];
  }
  return ret;
}
