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;
void _fill_int(int* p, int val, int rep) {
  int i;
  for (i = 0; i < rep; i++) p[i] = val;
}
signed long long GETi() {
  signed long long i;
  scanf("%lld", &i);
  return i;
}
int N, M;
int A[200100];
vector<int> E[200100];
int L[200100], R[200100], D[200100];
int id;
template <class V, int ME>
class BIT {
 public:
  V bit[1 << ME];
  V total(int entry) {
    V s = 0;
    entry++;
    while (entry > 0) s += bit[entry - 1], entry -= entry & -entry;
    return s;
  }
  void update(int entry, V val) {
    entry++;
    while (entry <= (1 << ME)) bit[entry - 1] += val, entry += entry & -entry;
  }
};
BIT<int, 20> bt[2];
void dfs(int cur, int pre) {
  int i, t = -1;
  L[cur] = ++id;
  for (i = 0; i < E[cur].size(); i++) {
    int tar = E[cur][i];
    if (tar == pre)
      t = i;
    else
      D[tar] = D[cur] + 1, dfs(tar, cur);
  }
  R[cur] = id;
}
void solve() {
  int f, i, j, k, l, x, y;
  cin >> N >> M;
  for (i = 0; i < N; i++) cin >> A[i];
  for (i = 0; i < N - 1; i++) {
    cin >> x >> y;
    E[x - 1].push_back(y - 1);
    E[y - 1].push_back(x - 1);
  }
  dfs(0, -1);
  for (i = 0; i < M; i++) {
    cin >> x;
    if (x == 1) {
      cin >> x >> y;
      bt[D[x - 1] & 1].update(L[x - 1], y);
      bt[D[x - 1] & 1].update(R[x - 1] + 1, -y);
    } else {
      cin >> x;
      y = bt[0].total(L[x - 1]) - bt[1].total(L[x - 1]);
      if (D[x - 1] & 1) y = -y;
      (void)printf("%d\n", A[x - 1] + y);
    }
  }
}
int main(int argc, char** argv) {
  string s;
  if (argc == 1) ios::sync_with_stdio(false);
  for (int i = 1; i < argc; i++) s += argv[i], s += '\n';
  for (int i = s.size() - 1; i >= 0; i--) ungetc(s[i], stdin);
  solve();
  return 0;
}
