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 long long int inf = 1e18;
const long long int mod = 1e9 + 7;
const long long int MOD = 998244353;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c *x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug &operator<<(const c &) {
    return *this;
  }
};
vector<char *> tokenizer(const char *args) {
  char *token = new char[111];
  strcpy(token, args);
  token = strtok(token, ", ");
  vector<char *> v({token});
  while ((token = strtok(NULL, ", "))) v.push_back(token);
  return reverse(v.begin(), v.end()), v;
}
#pragma clang diagnostic push
#pragma clang diagnostic ignored "-Wunused-parameter"
void debugg(vector<char *> args) { cerr << "\b\b "; }
#pragma clang diagnostic pop
template <typename Head, typename... Tail>
void debugg(vector<char *> args, Head H, Tail... T) {
  debug() << " [" << args.back() << ": " << H << "] ";
  args.pop_back();
  debugg(args, T...);
}
template <typename T>
T power(T a, T b) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2));
    if (b % 2)
      return (res * res * a);
    else
      return res * res;
  }
}
template <typename T>
T power(T a, T b, T modulo) {
  if (b == 0) return 1;
  if (b == 1)
    return a;
  else {
    T res = (power(a, b / 2, modulo) % modulo);
    if (b % 2)
      return ((((res % modulo) * (res % modulo)) % modulo) * (a % modulo)) %
             modulo;
    else
      return ((res % modulo) * (res % modulo)) % modulo;
  }
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
void factorial(vector<long long int> &fact, long long int n) {
  fact.resize(n + 1, 1);
  fact[0] = 1;
  fact[1] = 1;
  for (int i = 2; i <= n; ++i) {
    fact[i] = ((fact[i - 1] % mod) * (i % mod)) % mod;
  }
}
long long int mod_inv(long long int a) {
  return (power<long long int>(a, mod - 2, mod)) % mod;
}
long long int ncr(long long int n, long long int r,
                  vector<long long int> &fact) {
  if (r > n or n < 0 or r < 0) return 0LL;
  return (((fact[n] % mod) * (mod_inv(fact[n - r]) % mod)) % mod *
          (mod_inv(fact[r]) % mod)) %
         mod;
}
template <typename T>
class BIT {
  int n;
  vector<T> bit;

 public:
  BIT(int tree_size) {
    n = tree_size;
    bit.resize(n + 5);
  }
  bool update(int x, T val) {
    if (x > n or x < 1) return false;
    for (; x <= n; x += x & -x) {
      bit[x] += val;
    }
    return true;
  }
  bool range_update(int l, int r, T val) {
    update(l, val);
    update(r + 1, -val);
    return true;
  }
  T point_query(int x) {
    T sum = 0;
    for (; x > 0; x -= x & -x) sum += bit[x];
    return sum;
  }
};
const int N = 200009;
vector<int> start(N), finish(N), lvl(N);
vector<vector<long long int> > g;
int timer = 0;
void dfs(long long int node, long long int par, long long int level) {
  lvl[node] = level;
  start[node] = ++timer;
  for (auto &to : g[node]) {
    if (to == par) continue;
    dfs(to, node, level + 1);
  }
  finish[node] = timer;
}
int main() {
  long long int n, m;
  cin >> n >> m;
  vector<long long int> v(n + 1);
  for (auto i = 1; i <= n; i++) {
    cin >> v[i];
  };
  vector<long long int> a;
  g.assign(n + 1, a);
  for (int i = 1; i < n; ++i) {
    long long int x, y;
    cin >> x >> y;
    g[x].emplace_back(y);
    g[y].emplace_back(x);
  }
  dfs(1, 0, 0);
  {};
  {};
  {};
  BIT<int> bit(n);
  for (int i = 0; i < m; ++i) {
    int t;
    cin >> t;
    if (t == 1) {
      long long int x, val;
      cin >> x >> val;
      if (lvl[x] % 2) {
        val = -val;
      }
      bit.range_update(start[x], finish[x], val);
    } else {
      long long int x;
      cin >> x;
      long long int ans = bit.point_query(start[x]);
      if (lvl[x] & 1) ans = -ans;
      cout << ans + v[x] << "\n";
    }
  }
}
