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 err(istringstream &iss) {}
template <typename T, typename... Args>
void err(istringstream &iss, const T &varVal, const Args &...args) {
  string varName;
  iss >> varName;
  if (varName.back() == ',') varName.back() = ' ';
  cout << varName << " = " << varVal << "; ", err(iss, args...);
}
int n, m, K, T, Q, cn;
struct BIT {
  bool zeroIndxd = 0;
  static const int mxIdx = 200005 << 1;
  long long tree[mxIdx + 2] = {0};
  void addAtIdx(int idx, long long val) {
    idx += zeroIndxd;
    while (idx <= mxIdx) tree[idx] += val, idx += (idx & -idx);
  }
  long long getPrefixSum(int idx) {
    idx += zeroIndxd;
    long long sum = 0;
    while (idx > 0) sum += tree[idx], idx -= (idx & -idx);
    return sum;
  }
  int lower_bound(long long x) {
    int idx = 0, mask = 1 << (31 - __builtin_clz(200005 - 1)), ret = 200005;
    while (mask) {
      int tIdx = idx + mask;
      mask >>= 1;
      if (tIdx >= 200005 || tIdx <= 0) continue;
      if (tree[tIdx] >= x) ret = min(ret, tIdx);
      if (tree[tIdx] < x) idx = tIdx, x -= tree[tIdx];
    }
    ret -= zeroIndxd;
    return ret;
  }
  void multiplyAll(long long c) {
    for (int i = 1; i <= 200005; i++) tree[i] *= c;
  }
};
int a[200005], t, x, val, tiks, disc[200005], finish[200005], lvl[200005];
list<int> adj[200005];
void dfs(int u, int par, int dist) {
  lvl[u] = dist, disc[u] = ++tiks;
  for (auto &v : adj[u])
    if (v != par) {
      dfs(v, u, dist + 1);
    }
  finish[u] = ++tiks;
}
int main() {
  cin.tie(NULL);
  scanf("%d%d", &n, &Q);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v), adj[v].push_back(u);
  }
  dfs(1, -1, 0);
  BIT ft;
  while (Q--) {
    scanf("%d", &t);
    if (t == 1) {
      scanf("%d%d", &x, &val);
      if (lvl[x] & 1) val = -val;
      ft.addAtIdx(disc[x], val), ft.addAtIdx(finish[x], -val);
    } else
      scanf("%d", &x),
          printf("%lld\n",
                 ((lvl[x] & 1) ? -1 : 1) * ft.getPrefixSum(disc[x]) + a[x]);
  }
  return 0;
}
