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 MaxN = 200000 + 10;
int n, m, a[MaxN], l[MaxN], r[MaxN], deep[MaxN];
int bit[MaxN << 1], sl;
void add(int wz, int v) {
  for (int i = wz; i <= sl; i += ((i) & -(i))) bit[i] += v;
}
int qry(int wz) {
  int ret = 0;
  for (int i = wz; i > 0; i -= ((i) & -(i))) ret += bit[i];
  return ret;
}
int head[MaxN], nxt[MaxN << 1], to[MaxN << 1], top;
void addedge(int fr, int tt) {
  top++;
  nxt[top] = head[fr];
  to[top] = tt;
  head[fr] = top;
}
void init() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  int a, b;
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &a, &b);
    addedge(a, b);
    addedge(b, a);
  }
}
bool vis[MaxN];
void dfs(int x) {
  vis[x] = true;
  l[x] = ++sl;
  for (int i = head[x]; i; i = nxt[i])
    if (!vis[to[i]]) {
      deep[to[i]] = deep[x] + 1;
      dfs(to[i]);
    }
  r[x] = ++sl;
}
void solve() {
  deep[1] = 1;
  dfs(1);
  int op, x, v;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &op, &x);
    if (op == 1) {
      scanf("%d", &v);
      if (deep[x] & 1) v = -v;
      add(l[x], v);
      add(r[x], -v);
    } else {
      v = qry(l[x]);
      if (deep[x] & 1) v = -v;
      printf("%d\n", a[x] + v);
    }
  }
}
int main() {
  init();
  solve();
  return 0;
}
