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 MX = 212345;
int clk, vlu[MX], level[MX], st[MX], ed[MX];
int lazy_even[3 * MX][2], lazy_odd[3 * MX][2];
bool vis[MX];
vector<int> adjcnt[MX];
void dfs(int p, int u) {
  vis[u] = true;
  st[u] = ++clk;
  level[u] = level[p] + 1;
  for (int i = 0, sz = adjcnt[u].size(); i < sz; i++) {
    if (!vis[adjcnt[u][i]]) dfs(u, adjcnt[u][i]);
  }
  ed[u] = clk;
}
void propagate(int node, int lo, int hi, int tgl) {
  tgl == 1 ? lazy_odd[node][1] += lazy_odd[node][0]
           : lazy_even[node][1] += lazy_even[node][0];
  if (lo != hi) {
    int lft = node << 1, rgt = lft | 1;
    if (tgl == 1) {
      lazy_odd[lft][0] += lazy_odd[node][0];
      lazy_odd[rgt][0] += lazy_odd[node][0];
    } else {
      lazy_even[lft][0] += lazy_even[node][0];
      lazy_even[rgt][0] += lazy_even[node][0];
    }
  }
  tgl == 1 ? lazy_odd[node][0] = 0 : lazy_even[node][0] = 0;
}
void update(int node, int lo, int hi, int l, int r, int v, int tgl) {
  if (tgl == 1 && lazy_odd[node][0] != 0) {
    propagate(node, lo, hi, 1);
  }
  if (tgl == 2 && lazy_even[node][0] != 0) {
    propagate(node, lo, hi, 2);
  }
  if (lo > r || hi < l) return;
  if (lo >= l && hi <= r) {
    if (tgl == 1) {
      lazy_odd[node][0] += v;
      propagate(node, lo, hi, 1);
    } else {
      lazy_even[node][0] += v;
      propagate(node, lo, hi, 2);
    }
    return;
  }
  int mid = lo + (hi - lo) / 2, lft = node << 1, rgt = lft | 1;
  update(lft, lo, mid, l, r, v, tgl);
  update(rgt, mid + 1, hi, l, r, v, tgl);
}
int query(int node, int lo, int hi, int p, int tgl) {
  if (tgl == 1 && lazy_odd[node][0] != 0) {
    propagate(node, lo, hi, 1);
  }
  if (tgl == 2 && lazy_even[node][0] != 0) {
    propagate(node, lo, hi, 2);
  }
  if (lo == hi) {
    return (tgl == 1) ? lazy_odd[node][1] : lazy_even[node][1];
  }
  int mid = lo + (hi - lo) / 2, lft = node << 1, rgt = lft | 1;
  if (p <= mid)
    return query(lft, lo, mid, p, tgl);
  else
    return query(rgt, mid + 1, hi, p, tgl);
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", vlu + i);
  }
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d %d", &a, &b);
    adjcnt[a].push_back(b);
    adjcnt[b].push_back(a);
  }
  clk = 0, level[0] = 0;
  dfs(0, 1);
  while (m--) {
    int op, a, b;
    scanf("%d %d", &op, &a);
    if (op == 1) {
      scanf("%d", &b);
      if (level[a] & 1) {
        update(1, 1, clk, st[a], ed[a], b, 1);
        update(1, 1, clk, st[a], ed[a], -b, 2);
      } else {
        update(1, 1, clk, st[a], ed[a], b, 2);
        update(1, 1, clk, st[a], ed[a], -b, 1);
      }
    } else {
      int ans;
      ans = (level[a] & 1) ? query(1, 1, clk, st[a], 1)
                           : query(1, 1, clk, st[a], 2);
      printf("%d\n", ans + vlu[a]);
    }
  }
  return 0;
}
