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;
const int N = 1e6 + 5;
const int MOD = 1e9 + 7;
std::vector<long long int> v[N], euler;
bool visit[N];
long long int val[N], first[N], en[N], mx = N, bit[N], bit1[N], ht[N];
void dfs(long long int node, long long int hh) {
  visit[node] = 1;
  first[node] = (int)euler.size();
  ht[node] = hh;
  euler.push_back(node);
  for (auto i : v[node]) {
    if (!visit[i]) {
      dfs(i, hh + 1);
    }
  }
  en[node] = (int)euler.size();
}
void upd(long long int idx, long long int val) {
  while (idx < mx) {
    bit[idx] += val;
    idx += (idx & -idx);
  }
}
long long int get(long long int idx) {
  long long int sum = 0;
  while (idx > 0) {
    sum += bit[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
void upd1(long long int idx, long long int val) {
  while (idx < mx) {
    bit1[idx] += val;
    idx += (idx & -idx);
  }
}
long long int get1(long long int idx) {
  long long int sum = 0;
  while (idx > 0) {
    sum += bit1[idx];
    idx -= (idx & -idx);
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, q;
  cin >> n >> q;
  for (int i = 1; i <= n; i++) {
    cin >> val[i];
  }
  for (int i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  while (q--) {
    long long int type;
    cin >> type;
    if (type == 1) {
      long long int x, value;
      cin >> x >> value;
      if (ht[x] % 2 == 0) {
        upd(first[x] + 1, value);
        upd(en[x] + 1, -value);
      } else {
        upd1(first[x] + 1, value);
        upd1(en[x] + 1, -value);
      }
    } else {
      long long int x;
      cin >> x;
      long long int ans = val[x];
      if (ht[x] % 2 == 0) {
        ans += get(first[x] + 1);
        ans -= get1(first[x] + 1);
      } else {
        ans += get1(first[x] + 1);
        ans -= get(first[x] + 1);
      }
      cout << ans << "\n";
    }
  }
  return 0;
}
