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;
int val[300005], dep[300005];
int nxt[300005 * 2], vv[300005 * 2], p[300005];
int t1[300005], t2[300005], L[300005], R[300005], n, timer;
inline void update1(int x, int v) {
  while (x <= n) t1[x] += v, x += x & -x;
}
inline void update2(int x, int v) {
  while (x <= n) t2[x] += v, x += x & -x;
}
inline int read1(int x) {
  int ret = 0;
  while (x > 0) ret += t1[x], x -= x & -x;
  return ret;
}
inline int read2(int x) {
  int ret = 0;
  while (x > 0) ret += t2[x], x -= x & -x;
  return ret;
}
inline void dfs(int u, int fa) {
  L[u] = ++timer;
  for (int i = p[u]; i != -1; i = nxt[i]) {
    int v = vv[i];
    if (v == fa) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u);
  }
  R[u] = timer;
}
int main() {
  int m, u, v, idx = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &val[i]);
  memset(p, -1, sizeof(p));
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    nxt[idx] = p[u];
    vv[idx] = v, p[u] = idx++;
    nxt[idx] = p[v];
    vv[idx] = u, p[v] = idx++;
  }
  dep[1] = 1;
  timer = 0;
  dfs(1, 0);
  int add, op;
  while (m--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d", &u, &add);
      if (dep[u] & 1) {
        update1(L[u], add);
        update1(R[u] + 1, -add);
        update2(L[u], -add);
        update2(R[u] + 1, add);
      } else {
        update2(L[u], add);
        update2(R[u] + 1, -add);
        update1(L[u], -add);
        update1(R[u] + 1, add);
      }
    } else {
      scanf("%d", &u);
      printf("%d\n", val[u] + (dep[u] % 2 == 0 ? read2(L[u]) : read1(L[u])));
    }
  }
  return 0;
}
