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;
vector<int> adj[200001];
struct node {
  long long val;
  long long t[2];
} nodes[200001 << 2];
int n, m, sz, st[200001], fs[200001], h[200001], ver[200001], a[200001],
    b[200001];
void dfs(int v = 1, int p = 1, int hh = 0) {
  st[v] = ++sz;
  h[v] = hh;
  ver[sz] = v;
  for (auto u : adj[v]) {
    if (u != p) {
      dfs(u, v, hh + 1);
    }
  }
  fs[v] = sz;
}
int z(int i) { return i != 0 ? -1 : 1; }
void build(long long l = 1, long long r = n, long long id = 1) {
  if (l == r) {
    nodes[id].val = a[l];
    return;
  }
  long long mid = (l + r) / 2;
  build(l, mid, id * 2);
  build(mid + 1, r, id * 2 + 1);
}
void shift(long long l, long long r, long long id) {
  if (l == r) {
    nodes[id].val += z(1 - (h[ver[l]] % 2)) * nodes[id].t[1];
    nodes[id].val += z(0 - (h[ver[l]] % 2)) * nodes[id].t[0];
    nodes[id].t[1] = nodes[id].t[0] = 0;
  } else {
    nodes[id * 2].t[1] += nodes[id].t[1];
    nodes[id * 2].t[0] += nodes[id].t[0];
    nodes[id * 2 + 1].t[0] += nodes[id].t[0];
    nodes[id * 2 + 1].t[1] += nodes[id].t[1];
    nodes[id].t[1] = nodes[id].t[0] = 0;
  }
}
void updt(long long a, long long b, long long i, long long val, long long l = 1,
          long long r = n, long long id = 1) {
  shift(l, r, id);
  if (a > r || b < l) return;
  if (a <= l && b >= r) {
    nodes[id].t[i] += val;
    return;
  }
  long long mid = (l + r) / 2;
  updt(a, b, i, val, l, mid, id * 2);
  updt(a, b, i, val, mid + 1, r, id * 2 + 1);
}
long long get(long long pos, long long l = 1, long long r = n,
              long long id = 1) {
  shift(l, r, id);
  if (l == r) {
    return nodes[id].val;
  }
  long long mid = (l + r) / 2;
  if (pos <= mid) {
    return get(pos, l, mid, id * 2);
  } else {
    return get(pos, mid + 1, r, id * 2 + 1);
  }
}
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> b[i];
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs();
  for (int i = 1; i <= n; i++) {
    a[st[i]] = b[i];
  }
  build();
  for (int i = 1; i <= m; i++) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, val;
      cin >> x >> val;
      updt(st[x], fs[x], h[x] % 2, val);
    } else {
      long long x;
      cin >> x;
      cout << get(st[x]) << endl;
    }
  }
  return 0;
}
