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> A[200002];
int o[200002];
int B[200002];
int bp = 0;
int in[200002], out[200002];
int inoutp = 0;
int c1[200002], c2[200002];
bool s[200002];
int aib[400002];
void update(int pos, int val) {
  for (; pos < 400002; pos += pos & -pos) {
    aib[pos] += val;
  }
}
int query(int pos) {
  int sum = 0;
  for (; pos > 0; pos -= pos & -pos) {
    sum += aib[pos];
  }
  return sum;
}
void initc(int n) {
  o[0] = B[0] = 0;
  for (int i = 1; i <= n; i++) {
    c1[i] = c2[i] = (o[B[i]] - o[B[i - 1]]);
  }
}
void addb(int it) { B[++bp] = it; }
void dfs(int a, bool e) {
  in[a] = ++inoutp;
  s[a] = e;
  vector<int>::iterator it = A[a].begin();
  addb(a);
  while (it != A[a].end()) {
    dfs(*it, !e);
    it++;
  }
  out[a] = inoutp;
}
int ans(int n) {
  int res = 0;
  if (s[n]) {
    int k;
    for (k = 1; B[k] != n; k++) {
      res += c1[k];
    }
    res += c1[k];
  } else {
    int k;
    for (k = 1; B[k] != n; k++) {
      res += c2[k];
    }
    res += c2[k];
  }
  return res;
}
void printA(int n);
int main() {
  int n, m;
  int a, b;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &o[i]);
  }
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &a, &b);
    A[a].push_back(b);
  }
  dfs(1, 1);
  for (int i = 1, choice, val, pos; i <= m; i++) {
    scanf("%d", &choice);
    if (choice == 1) {
      scanf("%d%d", &pos, &val);
      int inp = in[pos];
      int outp = out[pos];
      if (s[pos]) {
        update(in[pos], val);
        update(out[pos] + 1, -val);
      } else {
        update(in[pos], -val);
        update(out[pos] + 1, val);
      }
    }
    if (choice == 2) {
      cin >> pos;
      if (s[pos])
        cout << o[pos] + query(in[pos]) << endl;
      else
        cout << o[pos] - query(in[pos]) << endl;
    }
  }
}
void printA(int n) {
  for (int i = 1; i <= n; i++) {
    cout << i << "*****";
    vector<int>::iterator it = A[i].begin();
    while (it != A[i].end()) {
      cout << *it << ' ';
      it++;
    }
    cout << endl;
  }
  cout << "!s[]= ";
  for (int i = 1; i <= n; i++) {
    cout << s[B[i]] << ' ';
  }
  cout << endl;
  cout << "B[]= ";
  for (int i = 1; i <= n; i++) {
    cout << B[i] << ' ';
  }
  cout << endl;
  cout << "in[]= ";
  for (int i = 1; i <= n; i++) {
    cout << in[i] << ' ';
  }
  cout << endl;
  cout << "out[]= ";
  for (int i = 1; i <= n; i++) {
    cout << out[i] << ' ';
  }
  cout << endl;
  cout << "s[]= ";
  for (int i = 1; i <= n; i++) {
    cout << s[i] << ' ';
  }
  cout << endl;
  cout << "c1[]= ";
  for (int i = 1; i <= n; i++) {
    cout << c1[i] << ' ';
  }
  cout << endl;
  cout << "c2[]= ";
  for (int i = 1; i <= n; i++) {
    cout << c2[i] << ' ';
  }
  cout << endl;
}
