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 TAM = 200000 + 10;
int n;
int a[TAM];
int b[2 * TAM];
int timer = 0;
int lvl[TAM];
int in[TAM];
int out[TAM];
vector<int> G[TAM];
int st[2][8 * TAM];
int lz[2][8 * TAM];
void dfs(int u, int pd) {
  in[u] = timer++;
  if (pd == -1)
    lvl[u] = 0;
  else
    lvl[u] = lvl[pd] + 1;
  for (int v : G[u]) {
    if (v != pd) dfs(v, u);
  }
  out[u] = timer++;
}
void build(int ind, int p, int l, int r) {
  if (l == r) {
    st[ind][p] = b[l];
    return;
  }
  int mid = (l + r) / 2;
  build(ind, 2 * p, l, mid);
  build(ind, 2 * p + 1, mid + 1, r);
}
void init() {
  dfs(0, -1);
  for (int i = 0; i < n; i++) {
    b[in[i]] = a[i];
    b[out[i]] = a[i];
  }
  build(0, 1, 0, 2 * n - 1);
  build(1, 1, 0, 2 * n - 1);
}
void push(int ind, int p, int l, int r) {
  if (lz[ind][p] != 0) {
    if (l == r) {
      st[ind][p] += lz[ind][p];
    } else {
      lz[ind][2 * p] += lz[ind][p];
      lz[ind][2 * p + 1] += lz[ind][p];
    }
    lz[ind][p] = 0;
  }
}
void update(int ind, int p, int l, int r, int tl, int tr, int val) {
  push(ind, p, l, r);
  if (r < tl or tr < l) return;
  if (tl <= l and r <= tr) {
    lz[ind][p] = val;
    push(ind, p, l, r);
    return;
  }
  int mid = (l + r) / 2;
  update(ind, 2 * p, l, mid, tl, tr, val);
  update(ind, 2 * p + 1, mid + 1, r, tl, tr, val);
}
int query(int ind, int p, int l, int r, int t) {
  push(ind, p, l, r);
  if (l == r) return st[ind][p];
  int mid = (l + r) / 2;
  if (t <= mid) return query(ind, 2 * p, l, mid, t);
  return query(ind, 2 * p + 1, mid + 1, r, t);
}
int main() {
  int q;
  scanf("%d %d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int u, v;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &u, &v);
    u--;
    v--;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  init();
  int op, val;
  for (int i = 0; i < q; i++) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d %d", &u, &val);
      u--;
      if (lvl[u] % 2 == 0) {
        update(0, 1, 0, 2 * n - 1, in[u], out[u], val);
        update(1, 1, 0, 2 * n - 1, in[u], out[u], -val);
      } else {
        update(0, 1, 0, 2 * n - 1, in[u], out[u], -val);
        update(1, 1, 0, 2 * n - 1, in[u], out[u], val);
      }
    } else {
      scanf("%d", &u);
      u--;
      if (lvl[u] % 2 == 0)
        printf("%d\n", query(0, 1, 0, 2 * n - 1, in[u]));
      else
        printf("%d\n", query(1, 1, 0, 2 * n - 1, in[u]));
    }
  }
  return 0;
}
