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 timer = 0, n, m;
vector<vector<int>> adj(200005 + 5);
vector<int> start(200005 + 5, 0);
vector<int> en(200005 + 5, 0);
vector<int> bit(200005 + 5, 0);
vector<int> depth(200005 + 5, 0);
void dfs(int u, int par) {
  int i, vector;
  start[u] = ++timer;
  for (i = 0; i < adj[u].size(); i++) {
    vector = adj[u][i];
    depth[vector] = depth[u] ^ 1;
    if (vector == par) continue;
    if (!start[vector]) dfs(vector, u);
  }
  en[u] = timer;
}
inline int query(int x) {
  int res = 0;
  while (x > 0) {
    res += bit[x];
    x = x & (x - 1);
  }
  return res;
}
inline void update(int x, int val) {
  while (x <= timer) {
    bit[x] += val;
    x += x & -x;
  }
}
int main() {
  int i, x, y, type, val;
  cin >> n >> m;
  vector<int> a(n + 1);
  for (i = 1; i <= n; i++) scanf(" %d", &a[i]);
  for (i = 1; i < n; i++)
    scanf(" %d %d", &x, &y), adj[x].push_back(y), adj[y].push_back(x);
  dfs(1, -1);
  while (m--) {
    scanf(" %d", &type);
    if (type == 1) {
      scanf(" %d %d", &x, &val);
      if (depth[x]) {
        update(en[x] + 1, val);
        update(start[x], -val);
      } else {
        update(start[x], val);
        update(en[x] + 1, -val);
      }
    } else {
      scanf(" %d", &x);
      if (depth[x])
        printf("%d\n", a[x] - query(start[x]));
      else
        printf("%d\n", a[x] + query(start[x]));
    }
  }
  return 0;
}
