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;
long long const N = 2e5 + 10;
long long const INF = 1e18;
long long const mod = 1000000009;
struct fnwk {
  fnwk(int n) : n(n + 5) { _BIT.resize(this->n, 0); }
  long long operator[](int i) {
    int sum = 0;
    i++;
    while (i > 0) {
      sum += _BIT[i];
      i -= i & (-i);
    }
    return sum;
  }
  void update(int i, int add) {
    i++;
    while (i < n) {
      _BIT[i] += add;
      i += i & (-i);
    }
  }

 private:
  int n;
  vector<int> _BIT;
};
long long n, q, v[N], vis[N], cntSubTree[N], myFather[N], myLeader[N],
    depthFromLeader[N], addedToLeader[N];
vector<int> e[N];
vector<pair<fnwk, fnwk>> leaderBITs;
map<int, int> chainSize;
int cntDfs(int idx) {
  vis[idx] = 1;
  int ret = 1;
  for (int to : e[idx])
    if (!vis[to]) {
      ret += cntDfs(to);
      myFather[to] = idx;
    }
  cntSubTree[idx] = ret;
  return ret;
}
void parting(int idx, int leader) {
  vis[idx] = 1;
  myLeader[idx] = leader;
  if (leader != idx) depthFromLeader[idx] = depthFromLeader[myFather[idx]] + 1;
  chainSize[leader]++;
  long long maxi = 0;
  for (int to : e[idx])
    if (!vis[to]) maxi = max(maxi, cntSubTree[to]);
  for (int to : e[idx])
    if (!vis[to]) {
      if (maxi == cntSubTree[to]) {
        parting(to, leader);
        maxi = 1e9;
      } else
        parting(to, to);
    }
  return;
}
long long getAns(int idx, int totDist, int leaderMemberDist) {
  if (idx == -1) return 0;
  long long ret = 0, tmp = 0;
  if (myLeader[idx] == idx) {
    if (leaderMemberDist % 2) {
      tmp = addedToLeader[idx] * -1;
      tmp += leaderBITs[idx].first[leaderMemberDist];
    } else {
      tmp = addedToLeader[idx];
      tmp += leaderBITs[idx].second[leaderMemberDist];
    }
    int tmpD = totDist - leaderMemberDist;
    if (tmpD % 2) tmp *= -1;
    ret += tmp + getAns(myFather[idx], totDist + 1, 0);
    return ret;
  } else {
    return getAns(myLeader[idx], totDist + depthFromLeader[idx],
                  depthFromLeader[idx]);
  }
  return ret;
}
int main() {
  cin >> n >> q;
  for (int i = 0; i < n; i++) scanf("%d", &v[i + 1]);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  cntDfs(1);
  memset(vis, 0, sizeof vis);
  parting(1, 1);
  myFather[1] = -1;
  leaderBITs.push_back(make_pair(fnwk(0), fnwk(0)));
  for (int i = 1; i <= n; i++) {
    leaderBITs.push_back(make_pair(fnwk(chainSize[i]), fnwk(chainSize[i])));
  }
  while (q--) {
    long long t, x, val;
    scanf("%lld%lld", &t, &x);
    if (t == 1) {
      scanf("%lld", &val);
      int ld = myLeader[x], dep = depthFromLeader[x];
      if (depthFromLeader[x] % 2) val *= -1;
      addedToLeader[ld] += val;
      if (dep % 2) {
        leaderBITs[ld].second.update(0, -val);
        leaderBITs[ld].second.update(dep, val);
        if (dep - 2 >= 0) {
          leaderBITs[ld].first.update(0, val);
          leaderBITs[ld].first.update(dep - 1, -val);
        }
      } else {
        if (dep - 1 >= 0) {
          leaderBITs[ld].first.update(0, val);
          leaderBITs[ld].first.update(dep, -val);
        }
        if (dep - 2 >= 0) {
          leaderBITs[ld].second.update(0, -val);
          leaderBITs[ld].second.update(dep - 1, val);
        }
      }
    } else {
      long long ans = getAns(x, 0, 0) + v[x];
      printf("%lld\n", ans);
    }
  }
  return 0;
}
