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 char IO_MODE = 0;
inline long long ReadInt() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline void WriteInt(long long x) {
  char CH[20];
  int Num = 0;
  if (!x) {
    putchar('0');
    return;
  }
  while (x > 0) CH[++Num] = x % 10, x /= 10;
  while (Num) putchar(CH[Num--] + 48);
}
inline void out(int x) {
  if (IO_MODE & 1)
    WriteInt(x);
  else
    printf("%i", x);
}
template <typename... Args>
inline void out(int x, Args... args) {
  out(x);
  putc(' ', stdout);
  out(args...);
}
inline void in(int &x) {
  if (IO_MODE & 2)
    x = ReadInt();
  else
    scanf("%i", &x);
}
template <typename... Args>
inline void in(int &x, Args &...args) {
  in(x);
  in(args...);
}
inline void outll(long long x) {
  if (IO_MODE & 1)
    WriteInt(x);
  else
    printf("%I64d", x);
}
template <typename... Args>
inline void outll(long long x, Args... args) {
  outll(x);
  putc(' ', stdout);
  outll(args...);
}
inline void inll(long long &x) {
  if (IO_MODE & 2)
    x = ReadInt();
  else
    scanf("%I64d", &x);
}
template <typename... Args>
inline void inll(long long &x, Args &...args) {
  inll(x);
  inll(args...);
}
int aa, bb, cc, dd, ee, ff, gg, hh, ii, jj, kk, mm, nn, oo, pp, qq, rr, ss, tt,
    uu, vv, ww, xx, yy, zz;
int tc, ty;
int n, m;
int A[200100];
int In[200100], Out[200100];
int Lev[200100];
vector<vector<int>> adjList;
struct fenwick {
  int n;
  vector<int> fn;
  fenwick(int sz = 200100) {
    n = sz;
    fn.assign(n + 2, 0);
  }
  void upd(int L, int R, int val) {
    for (int i = L; i <= n; i += i & -i) fn[i] += val;
    for (int i = R + 1; i <= n; i += i & -i) fn[i] -= val;
  }
  int qry(int i) {
    int ret = 0;
    for (; i; i -= i & -i) ret += fn[i];
    return ret;
  }
} fn[2];
int cnt;
void dfs(int u, int p, int lev) {
  Lev[u] = lev;
  In[u] = ++cnt;
  for (auto v : adjList[u])
    if (v != p) dfs(v, u, !lev);
  Out[u] = cnt;
}
int main() {
  in(n, m);
  adjList.resize(n + 1);
  for (int i = 1; i <= n; i++) in(A[i]);
  for (int i = 0; i < n - 1; i++) {
    in(uu, vv);
    adjList[uu].push_back(vv);
    adjList[vv].push_back(uu);
  }
  dfs(1, -1, 0);
  int u, val;
  while (m--) {
    if (ReadInt() & 1) {
      in(u, val);
      fn[Lev[u]].upd(In[u], Out[u], val);
      fn[!Lev[u]].upd(In[u], Out[u], -val);
    } else {
      in(u);
      out(fn[Lev[u]].qry(In[u]) + A[u]);
      putc('\n', stdout);
    }
  }
}
