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>
const long long int M = 1e9 + 7;
using namespace std;
long long int bit[411005] = {0};
long long int a[200004];
vector<int> g[200004];
long long int lev[200004];
long long int tin[200004];
long long int tout[200004];
int v = 1;
void dfs(int s, int par) {
  tin[s] = v++;
  lev[s] = lev[par] + 1;
  for (int x : g[s]) {
    if (x != par) dfs(x, s);
  }
  tout[s] = v++;
}
void update(int idx, long long int val) {
  while (idx <= 401005) {
    bit[idx] += val;
    idx += (idx & -idx);
  }
}
long long int query(int idx) {
  long long int ans = 0;
  while (idx > 0) {
    ans += bit[idx];
    idx -= (idx & -idx);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n, m, q, x, y;
  ;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  lev[0] = 1;
  dfs(1, 0);
  while (m--) {
    int type;
    cin >> type;
    if (type == 1) {
      cin >> x >> y;
      if (lev[x] & 1) {
        update(tin[x], -y);
        update(tout[x], y);
      } else {
        update(tin[x], y);
        update(tout[x], -y);
      }
    } else {
      cin >> x;
      if (lev[x] & 1)
        cout << a[x] - query(tin[x]) << endl;
      else
        cout << a[x] + query(tin[x]) << endl;
    }
  }
  return 0;
}
