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;
int rd(int &x) { return scanf("%d", &x); }
int rd(int &x, int &y) { return scanf("%d %d", &x, &y); }
int rd(int &x, int &y, int &z) { return scanf("%d %d %d", &x, &y, &z); }
int rd(int &x, int &y, int &z, int &s) {
  return scanf("%d %d %d %d", &x, &y, &z, &s);
}
long long rd(long long &x) { return scanf("%lld", &x); }
long long rd(long long &x, long long &y) { return scanf("%lld %lld", &x, &y); }
long long rd(long long &x, long long &y, long long &z) {
  return scanf("%lld %lld %lld", &x, &y, &z);
}
long long rd(long long &x, long long &y, long long &z, long long &s) {
  return scanf("%lld %lld %lld %lld", &x, &y, &z, &s);
}
struct debugger {
  template <class T>
  debugger &operator,(const T &v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
template <typename T>
void debugv(vector<T> vec) {
  cerr << "printing vector:: ";
  for (int i = 0; i < (vec.size()); i++) {
    cout << vec[i] << " ";
  }
  cerr << endl;
}
template <typename T>
void debugs(set<T> s) {
  cerr << "printing set:: ";
  for (auto it = s.begin(); it != s.end(); it++) {
    cout << (*it) << " ";
  }
  cerr << endl;
}
const long long N = 2e5 + 8;
int t[2][N], a[N], n, q, in[N], out[N], c;
vector<int> g[N];
bool b[N];
void dfs(int x, int p, bool d) {
  in[x] = ++c;
  b[x] = d;
  for (int i = 0, sz = g[x].size(); i < sz; i++) {
    int y = g[x][i];
    if (p == y) continue;
    dfs(y, x, d ^ 1);
  }
  out[x] = c;
}
void update(int id, int x, int y, int n) {
  while (id <= n) {
    t[y][id] += x;
    id += id & (-id);
  }
}
long long query(int id, int y) {
  long long sum = 0;
  while (id > 0) {
    sum += t[y][id];
    id -= id & (-id);
  }
  return sum;
}
int main() {
  rd(n, q);
  for (int i = 1; i <= (n); i++) rd(a[i]);
  for (int i = 1; i <= (n - 1); i++) {
    int u, v;
    rd(u, v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, -1, 0);
  for (int i = 1; i <= (q); i++) {
    long long tp, x, val;
    rd(tp, x);
    if (tp == 1) {
      rd(val);
      update(in[x], val, b[x], 2e5 + 6);
      update(out[x] + 1, -val, b[x], 2e5 + 6);
    } else {
      long long p = query(in[x], b[x]);
      long long q = query(in[x], b[x] ^ 1);
      printf("%lld\n", a[x] + p - q);
    }
  }
  return 0;
}
