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 a[2000005], aa, b, i, n, q, v1, t, fir[2000005], las[2000005],
    level[2000005], bit[1000005];
vector<long long> v[2000005];
vector<long long> euler;
void dfs(long long x, long long pa, long long lev) {
  level[x] = lev;
  fir[x] = euler.size();
  euler.push_back(x);
  for (long long i = 0; i < v[x].size(); i++) {
    if (v[x][i] == pa) continue;
    dfs(v[x][i], x, lev + 1);
  }
  las[x] = euler.size();
  euler.push_back(x);
}
void update(long long idx, long long val) {
  while (idx <= 2 * n) {
    bit[idx] += val;
    idx += idx & -idx;
  }
}
long long read(long long idx) {
  long long sum = 0;
  while (idx > 0) {
    sum += bit[idx];
    idx -= idx & -idx;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  euler.push_back(0);
  cin >> n;
  cin >> q;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i < n; i++) {
    cin >> aa >> b;
    v[aa].push_back(b);
    v[b].push_back(aa);
  }
  dfs(1, 0, 0);
  long long x;
  for (i = 1; i <= q; i++) {
    cin >> t >> v1;
    if (t == 1) {
      cin >> x;
      if (level[v1] % 2 == 0) {
        update(fir[v1], x);
        update(las[v1] + 1, -x);
      } else {
        update(fir[v1], -x);
        update(las[v1] + 1, x);
      }
    } else if (t == 2) {
      if (level[v1] % 2 == 0) {
        cout << a[v1] + read(fir[v1]);
      } else {
        cout << a[v1] - read(fir[v1]);
      }
      cout << "\n";
    }
  }
  return 0;
}
