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;
template <class T>
inline void umax(T &x, T y) {
  if (y > x) x = y;
}
template <class T>
inline void umin(T &x, T y) {
  if (y < x) x = y;
}
const int N = 2e5 + 5, B = 0x7fffffff;
int in[N], out[N], lvl[N], now[N];
int st[4 * N], ger[N];
int n, m, tim;
vector<int> g[N];
void dfs(int nd, int par) {
  in[nd] = ++tim;
  ger[tim] = now[nd];
  lvl[nd] = lvl[par] + 1;
  for (auto adj : g[nd])
    if (adj != par) dfs(adj, nd);
  out[nd] = tim;
}
void upd(int l, int r, int id, int ql, int qr, int x) {
  if (r < ql || l > qr) return;
  if (l >= ql && r <= qr) {
    st[id] += x;
    return;
  }
  int mid = (l + r) >> 1;
  upd(l, mid, id << 1, ql, qr, x);
  upd(mid + 1, r, id << 1 | 1, ql, qr, x);
}
int get(int l, int r, int id, int u) {
  if (l == r) return st[id];
  int mid = (l + r) >> 1;
  if (u <= mid)
    return st[id] + get(l, mid, id << 1, u);
  else
    return st[id] + get(mid + 1, r, id << 1 | 1, u);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &now[i]);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= m; i++) {
    int ty, no, x;
    scanf("%d", &ty);
    if (ty == 1) {
      scanf("%d%d", &no, &x);
      if (lvl[no] % 2 == 0)
        upd(1, n, 1, in[no], out[no], x);
      else
        upd(1, n, 1, in[no], out[no], -x);
    } else {
      scanf("%d", &no);
      if (lvl[no] % 2 == 0)
        printf("%d\n", ger[in[no]] + get(1, n, 1, in[no]));
      else
        printf("%d\n", ger[in[no]] - get(1, n, 1, in[no]));
    }
  }
  return !1;
}
