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 inf = 1e9;
const long long int INF = 1e18;
const int N = 100005;
const int MOD = 1e9 + 7;
const double EPS = 1e-6;
const double PI = acos(-1.0);
long long int n, m;
long long int ara[2 * N];
long long int Start[2 * N], Finish[2 * N], Level[2 * N];
long long int Time = 0;
long long int Tree[2 * 4 * N], Lazy[2 * 4 * N];
vector<int> graph[2 * N];
void Propagate(long long int node, long long int b, long long int e) {
  if (Lazy[node] != 0) {
    Tree[node] += (Lazy[node]);
    if (b != e) {
      Lazy[2 * node] += Lazy[node];
      Lazy[2 * node + 1] += Lazy[node];
    }
    Lazy[node] = 0;
  }
}
void Update(long long int node, long long int b, long long int e,
            long long int i, long long int j, long long int val) {
  Propagate(node, b, e);
  if (i > e || j < b) return;
  if (i <= b && j >= e) {
    Tree[node] += val;
    if (b != e) {
      Lazy[2 * node] += val;
      Lazy[2 * node + 1] += val;
    }
    return;
  }
  long long int left = 2 * node, right = 2 * node + 1, mid = (b + e) / 2;
  Update(left, b, mid, i, j, val);
  Update(right, mid + 1, e, i, j, val);
  Tree[node] = Tree[left] + Tree[right];
}
long long int Query(long long int node, long long int b, long long int e,
                    long long int pos) {
  Propagate(node, b, e);
  if (pos > e || pos < b) return 0LL;
  if (b == e && b == pos) return Tree[node];
  long long int left = 2 * node, right = 2 * node + 1, mid = (b + e) / 2;
  long long int q1 = Query(left, b, mid, pos);
  long long int q2 = Query(right, mid + 1, e, pos);
  return (q1 + q2);
}
void dfs(int s, int p) {
  Start[s] = ++Time;
  Level[s] = Level[p] + 1;
  for (int i = 0; i < graph[s].size(); i++) {
    int next = graph[s][i];
    if (next == p) continue;
    dfs(next, s);
  }
  Finish[s] = Time;
}
int main(int argc, char const *argv[]) {
  scanf("%lld %lld", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &ara[i]);
  }
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
  Time = 0;
  dfs(1, 0);
  while (m--) {
    int op;
    scanf("%d", &op);
    if (op == 1) {
      int pos, val;
      scanf("%d %d", &pos, &val);
      int from = Start[pos], to = Finish[pos];
      if (Level[pos] & 1)
        Update(1, 1, 2 * N, from, to, +val);
      else
        Update(1, 1, 2 * N, from, to, -val);
    } else {
      int pos;
      scanf("%d", &pos);
      int x = Start[pos];
      long long int ans = Query(1, 1, 2 * N, x);
      if (Level[pos] & 1)
        printf("%lld\n", ara[pos] + ans);
      else
        printf("%lld\n", ara[pos] - ans);
    }
  }
  return 0;
}
