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;
using namespace std;
long long powm(long long base, long long exp, long long mod = 1000000007) {
  long long ans = 1;
  while (exp) {
    if (exp & 1) ans = (ans * base) % mod;
    exp >>= 1, base = (base * base) % mod;
  }
  return ans;
}
long long n, u, v, a[200005], st[2][200005], en[2][200005], timer[2],
    tour[2][200005], level[200005], parent[200005], ft[2][200005];
vector<long long> adj[200005];
void dfs(long long mode, long long curr, long long par, long long now) {
  parent[curr] = par, level[curr] = now;
  if (now == mode)
    st[mode][curr] = (++timer[mode]), tour[mode][timer[mode]] = curr;
  for (auto i : adj[curr])
    if (i != par) dfs(mode, i, curr, now ^ 1);
  if (now == mode) en[mode][curr] = timer[mode];
}
void update(long long mode, long long p, long long val) {
  long long curr = p;
  for (; curr <= 200005; curr += (curr & (-curr))) ft[mode][curr] += val;
}
void update(long long mode, long long a, long long b, long long val) {
  update(mode, a, val), update(mode, b + 1, -val);
}
long long query(long long mode, long long idx) {
  long long sum = 0, curr = idx;
  for (; curr > 0; curr -= (curr & (-curr))) sum += ft[mode][curr];
  return sum;
}
int main() {
  long long type, x, val, m;
  cin >> n >> m;
  for (long long i = 1; i < n + 1; i++) cin >> a[i];
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v), adj[v].push_back(u);
  }
  for (long long i = 0; i < 2; i++) dfs(i, 1, 0, 0);
  while (m--) {
    cin >> type;
    if (type == 1) {
      cin >> x >> val;
      long long mode = level[x], tmp = adj[x].size();
      long long p = st[mode][x], q = en[mode][x], first = -1, last = -1;
      update(mode, p, q, val);
      for (auto i : adj[x])
        if (i != parent[x]) {
          first = i;
          break;
        }
      for (long long i = tmp - 1; i >= 0; i--)
        if (adj[x][i] != parent[x]) {
          last = adj[x][i];
          break;
        }
      if (first != -1) {
        mode ^= 1;
        p = st[mode][first], q = en[mode][last];
        update(mode, p, q, -val);
      }
    } else {
      cin >> x;
      cout << query(level[x], st[level[x]][x]) + a[x] << endl;
    }
  }
  return 0;
}
