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;
int n, q, v[200010], ini[200010], fim[200010], cnt, deep[200010];
vector<int> grafo[200010];
void dfs(int x, int p) {
  deep[x] = deep[p] + 1;
  ini[x] = ++cnt;
  for (auto v : grafo[x]) {
    if (v == p) continue;
    dfs(v, x);
  }
  fim[x] = cnt;
}
struct seg {
  int tree[4 * 200010], lazy[4 * 200010];
  inline void prop(int nod, int a, int b) {
    if (!lazy[nod]) return;
    tree[nod] += (b - a + 1) * lazy[nod];
    if (a != b) {
      lazy[(2 * nod)] += lazy[nod];
      lazy[(2 * nod + 1)] += lazy[nod];
    }
    lazy[nod] = 0;
  }
  void upd(int nod, int a, int b, int i, int j, int x) {
    prop(nod, a, b);
    if (j < a or i > b) return;
    if (i <= a and j >= b) {
      tree[nod] += (b - a + 1) * x;
      if (a != b) {
        lazy[(2 * nod)] += x;
        lazy[(2 * nod + 1)] += x;
      }
      return;
    }
    upd((2 * nod), a, ((a + b) / 2), i, j, x),
        upd((2 * nod + 1), ((a + b) / 2) + 1, b, i, j, x);
    tree[nod] = tree[(2 * nod)] + tree[(2 * nod + 1)];
  }
  int query(int nod, int a, int b, int i, int j) {
    prop(nod, a, b);
    if (j < a or i > b) return 0;
    if (i <= a and j >= b) return tree[nod];
    return query((2 * nod), a, ((a + b) / 2), i, j) +
           query((2 * nod + 1), ((a + b) / 2) + 1, b, i, j);
  }
} T[2];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> q;
  for (int i = 1; i <= n; i++) cin >> v[i];
  for (int i = 1, a, b; i < n; i++) {
    cin >> a >> b;
    grafo[a].push_back(b);
    grafo[b].push_back(a);
  }
  dfs(1, 1);
  for (int i = 1, a, b, c; i <= q; i++) {
    cin >> a >> b;
    if (a == 1) {
      cin >> c;
      T[deep[b] % 2].upd(1, 1, n, ini[b], fim[b], c);
    } else {
      int A = T[deep[b] % 2].query(1, 1, n, ini[b], ini[b]);
      int B = T[(deep[b] + 1) % 2].query(1, 1, n, ini[b], ini[b]);
      cout << A - B + v[b] << "\n";
    }
  }
}
