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;
long long t, n, u, v, m, q, r, ql, qr, k, d, l, w, c, x;
const int N = 2e5 + 500;
const long long mod = 1e9 + 7;
const long long INF = 1LL << 57LL;
const bool JUDGE = false;
int T[2][N];
int a[N];
int pos[N], depth[N], HB[N];
vector<int> adj[N];
int sz[N];
int HA = 0;
void modify(int id, int i, int v) {
  while (i <= n) {
    T[id][i] += v;
    i += (i & -i);
  }
}
int get(int id, int i) {
  int ret = 0;
  while (i > 0) {
    ret += T[id][i];
    i -= (i & -i);
  }
  return ret;
}
void dfs(int v, int p, int dep) {
  depth[v] = dep;
  pos[v] = ++HA;
  HB[HA] = v;
  sz[v] = 1;
  for (long long i = 0; i < adj[v].size(); i++) {
    int u = adj[v][i];
    if (u == p) continue;
    dfs(u, v, dep + 1);
    sz[v] += sz[u];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  if (JUDGE) {
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
  }
  cin >> n >> q;
  for (long long i = 0; i < n; i++) cin >> a[i + 1];
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1, 0);
  for (long long i = 1; i < n + 1; i++) {
    int v = HB[i];
    modify((depth[v] & 1), pos[v], a[v]);
    modify((depth[v] & 1), pos[v] + 1, -a[v]);
  }
  while (q--) {
    int id;
    cin >> id;
    if (id == 1) {
      cin >> v >> x;
      int g = depth[v] & 1;
      int f = g ^ 1;
      modify(g, pos[v], x);
      modify(g, pos[v] + sz[v], -x);
      modify(f, pos[v], -x);
      modify(f, pos[v] + sz[v], x);
    } else {
      cin >> v;
      int f = depth[v] & 1;
      cout << get(f, pos[v]) << '\n';
    }
  }
  return 0;
}
