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 sq = 3000;
const int maxn = 2e5 + 10;
long long int a[maxn];
vector<pair<int, int>> v;
vector<int> adj[maxn];
int startingtime[maxn];
int finishingtime[maxn];
int h[maxn];
long long int b[maxn];
long long int c[maxn];
int n;
int m;
int cnt;
void nfa(int s, int pr) {
  a[s] += b[s];
  a[s] -= c[s];
  for (int i = 0; i < adj[s].size(); i++) {
    if (adj[s][i] != pr) {
      b[adj[s][i]] += c[s];
      c[adj[s][i]] += b[s];
      nfa(adj[s][i], s);
    }
  }
  b[s] = 0;
  c[s] = 0;
}
void upd() {
  for (int i = 0; i < v.size(); i++) {
    b[v[i].first] += v[i].second;
  }
  v.clear();
  nfa(1, -1);
}
void dfs(int s, int pr) {
  startingtime[s] = cnt;
  cnt++;
  for (int i = 0; i < adj[s].size(); i++) {
    if (adj[s][i] != pr) {
      dfs(adj[s][i], s);
    }
  }
  finishingtime[s] = cnt;
  return;
}
void dfa(int s, int pr) {
  for (int i = 0; i < adj[s].size(); i++) {
    if (adj[s][i] != pr) {
      h[adj[s][i]] = h[s] + 1;
      dfa(adj[s][i], s);
    }
  }
  return;
}
int main() {
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 1; i < n; i++) {
    int l;
    int r;
    scanf("%d", &l);
    scanf("%d", &r);
    adj[l].push_back(r);
    adj[r].push_back(l);
  }
  dfs(1, -1);
  h[1] = 1;
  dfa(1, -1);
  for (int i = 0; i < m; i++) {
    int g;
    scanf("%d", &g);
    if (g == 2) {
      int l;
      scanf("%d", &l);
      long long int f = a[l];
      for (int j = 0; j < v.size(); j++) {
        if (startingtime[v[j].first] <= startingtime[l] &&
            finishingtime[v[j].first] >= finishingtime[l]) {
          if ((h[l] - h[v[j].first]) % 2 == 1) {
            f -= v[j].second;
          } else {
            f += v[j].second;
          }
        }
      }
      printf("%d\n", f);
    } else {
      int l;
      int r;
      scanf("%d", &l);
      scanf("%d", &r);
      v.push_back({l, r});
      if (v.size() == sq) {
        upd();
      }
    }
  }
  return 0;
}
