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 dx[] = {-1, 0, 1, 0};
const int dy[] = {0, -1, 0, 1};
const int N = 2e5 + 4;
const long long MOD = 1e9 + 7;
long long INF = 2e18;
long long power(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  while (b) {
    if (b & 1LL) {
      res *= a;
      res %= mod;
    }
    a = a * a % mod;
    b >>= 1LL;
  }
  return res % mod;
}
int n;
void sos(int l, int r, int arr[]) {
  if (l + 1 == r) return;
  int mid = (l + r) >> 1;
  sos(l, mid, arr);
  sos(mid, r, arr);
  int len = (r - l) >> 1;
  for (int i = l; i < mid; i++) {
    arr[i + len] += arr[i];
  }
}
vector<vector<int> > g(N, vector<int>());
int lev[N], timer = 0, start[N], finish[N];
int BIT[N];
void upd(int x, int val) {
  for (; x < N; x += x & -x) BIT[x] += val;
}
int qry(int x) {
  int ans = 0;
  for (; x > 0; x -= x & -x) ans += BIT[x];
  return ans;
}
void dfs(int curr, int par, int l) {
  start[curr] = ++timer;
  lev[curr] = l;
  for (int x : g[curr]) {
    if (x != par) dfs(x, curr, l ^ 1);
  }
  finish[curr] = timer;
}
int main() {
  int m;
  cin >> n >> m;
  int u, v, a[n + 4];
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0, 0);
  while (m--) {
    int ty;
    cin >> ty;
    if (ty == 1) {
      int x, val;
      cin >> x >> val;
      if (lev[x]) {
        upd(start[x], -val);
        upd(finish[x] + 1, val);
      } else {
        upd(start[x], val);
        upd(finish[x] + 1, -val);
      }
    } else {
      int x;
      cin >> x;
      if (lev[x])
        cout << a[x] - qry(start[x]) << '\n';
      else
        cout << a[x] + qry(start[x]) << '\n';
    }
  }
  return 0;
}
