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;
const int MAX_N = 200010;
const int MAX_E = 2 * MAX_N;
int adj[MAX_N];
int ant[MAX_E], to[MAX_E], z = 0;
inline void add(int a, int b) {
  to[z] = b;
  ant[z] = adj[a];
  adj[a] = z++;
}
int A[MAX_N];
int bit[MAX_N], n;
void update(int x, int v) {
  while (x <= n) {
    bit[x] += v;
    x += x & -x;
  }
}
int sum(int x) {
  int ret = 0;
  while (x > 0) {
    ret += bit[x];
    x -= x & -x;
  }
  return ret;
}
int in[MAX_N], out[MAX_N], level[MAX_N], tempo = 1;
bool mark[MAX_N];
int dfs(int u) {
  in[u] = tempo;
  out[u] = tempo++;
  for (int i = adj[u]; ~i; i = ant[i]) {
    if (!mark[to[i]]) {
      level[to[i]] = 1 + level[u];
      mark[to[i]] = 1;
      out[u] = max(out[u], dfs(to[i]));
    }
  }
  return out[u];
}
int main() {
  cin.sync_with_stdio(false);
  memset((adj), (-1), sizeof(adj));
  memset((mark), (0), sizeof(mark));
  z = 0;
  tempo = 1;
  int m;
  cin >> n >> m;
  for (int i = (1); i < (n + 1); ++i) cin >> A[i];
  for (int i = (0); i < (n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    add(u, v);
    add(v, u);
  }
  mark[1] = 1;
  level[1] = 0;
  dfs(1);
  memset((bit), (0), sizeof(bit));
  while (m--) {
    int op;
    cin >> op;
    if (op == 1) {
      int u, v;
      cin >> u >> v;
      if (level[u] % 2) {
        update(in[u], v);
        update(out[u] + 1, -v);
      } else {
        update(in[u], -v);
        update(out[u] + 1, v);
      }
    } else {
      int u;
      cin >> u;
      if (level[u] % 2) {
        cout << A[u] + sum(in[u]) << endl;
      } else {
        cout << A[u] - sum(in[u]) << endl;
      }
    }
  }
  return 0;
}
