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;
struct node {
  int l, r, val;
  node *lc, *rc;
};
node *makeNode(int l, int r) {
  node *R = (node *)malloc(sizeof(node));
  R->l = l, R->r = r;
  R->lc = R->rc = NULL;
  R->val = 0;
  return R;
}
int N, M, A[200005], D[200005], T[200005], lo[200005], hi[200005], H;
vector<int> edge[200005];
bool vst[200005];
node *Even, *Odd;
void traversal(int node) {
  vst[node] = true;
  T[++H] = node;
  lo[node] = H;
  for (int i = 0; i < edge[node].size(); i++) {
    int nxt = edge[node][i];
    if (vst[nxt]) continue;
    D[nxt] = D[node] + 1;
    traversal(nxt);
  }
  hi[node] = H;
}
void maketree(node *cur) {
  if (cur->l == cur->r) return;
  int mid = (cur->l + cur->r) / 2;
  cur->lc = makeNode(cur->l, mid);
  cur->rc = makeNode(mid + 1, cur->r);
  maketree(cur->lc);
  maketree(cur->rc);
}
void update(node *cur, int left, int right, int val) {
  if (left <= cur->l && cur->r <= right) {
    cur->val += val;
    return;
  }
  int mid = (cur->l + cur->r) / 2;
  if (left <= mid) update(cur->lc, left, right, val);
  if (right > mid) update(cur->rc, left, right, val);
}
int answer(node *cur, int index) {
  int mid = (cur->l + cur->r) / 2;
  if (cur->l == cur->r) return cur->val;
  if (index <= mid)
    return answer(cur->lc, index) + cur->val;
  else
    return answer(cur->rc, index) + cur->val;
}
int main(void) {
  scanf("%d %d", &N, &M);
  for (int i = 1; i <= N; i++) scanf("%d", &A[i]);
  for (int i = 1; i < N; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    edge[v].push_back(u);
    edge[u].push_back(v);
  }
  D[1] = 1;
  traversal(1);
  Even = makeNode(1, N);
  Odd = makeNode(1, N);
  maketree(Even);
  maketree(Odd);
  for (int i = 1; i <= M; i++) {
    int query, x;
    scanf("%d %d", &query, &x);
    if (query == 1) {
      int val;
      scanf("%d", &val);
      A[x] += val;
      if (lo[x] < hi[x]) {
        int vl = val;
        if (D[x] & 1) vl *= -1;
        update(Even, lo[x] + 1, hi[x], vl);
        update(Odd, lo[x] + 1, hi[x], -vl);
      }
    } else {
      int add = (D[x] & 1) ? answer(Odd, lo[x]) : answer(Even, lo[x]);
      printf("%d\n", A[x] + add);
    }
  }
  return 0;
}
