Problem: There are n cities in Cyberland, numbered from 1 to n, connected by m bidirectional roads. The j-th road connects city aj and bj.

For tourists, souvenirs are sold in every city of Cyberland. In particular, city i sell it at a price of wi.

Now there are q queries for you to handle. There are two types of queries:

  * "C a w": The price in city a is changed to w.
  * "A a b": Now a tourist will travel from city a to b. He will choose a route, he also doesn't want to visit a city twice. He will buy souvenirs at the city where the souvenirs are the cheapest (possibly exactly at city a or b). You should output the minimum possible price that he can buy the souvenirs during his travel.



More formally, we can define routes as follow:

  * A route is a sequence of cities [x1, x2, ..., xk], where k is a certain positive integer.
  * For any 1 ≤ i < j ≤ k, xi ≠ xj.
  * For any 1 ≤ i < k, there is a road connecting xi and xi + 1.
  * The minimum price of the route is min(wx1, wx2, ..., wxk).
  * The required answer is the minimum value of the minimum prices of all valid routes from a to b.

Input

The first line of input contains three integers n, m, q (1 ≤ n, m, q ≤ 105), separated by a single space.

Next n lines contain integers wi (1 ≤ wi ≤ 109).

Next m lines contain pairs of space-separated integers aj and bj (1 ≤ aj, bj ≤ n, aj ≠ bj).

It is guaranteed that there is at most one road connecting the same pair of cities. There is always at least one valid route between any two cities.

Next q lines each describe a query. The format is "C a w" or "A a b" (1 ≤ a, b ≤ n, 1 ≤ w ≤ 109).

Output

For each query of type "A", output the corresponding answer.

Examples

Input

3 3 3
1
2
3
1 2
2 3
1 3
A 2 3
C 1 5
A 2 3


Output

1
2


Input

7 9 4
1
2
3
4
5
6
7
1 2
2 5
1 5
2 3
3 4
2 4
5 6
6 7
5 7
A 2 3
A 6 4
A 6 7
A 3 3


Output

2
1
5
3

Note

For the second sample, an optimal routes are:

From 2 to 3 it is [2, 3].

From 6 to 4 it is [6, 5, 1, 2, 4].

From 6 to 7 it is [6, 5, 7].

From 3 to 3 it is [3].

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void write(int x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
int n, m, Q, w[100004];
int p[200004], tms, cnt, low[100004], vis[100004];
vector<int> g[100004], G[200004];
stack<int> st;
void tarjan(int x) {
  vis[x] = low[x] = ++tms;
  st.push(x);
  for (int i = 0; i < g[x].size(); i++) {
    if (vis[g[x][i]])
      low[x] = min(low[x], vis[g[x][i]]);
    else {
      tarjan(g[x][i]), low[x] = min(low[x], low[g[x][i]]);
      if (low[g[x][i]] == vis[x]) {
        ++cnt;
        p[cnt] = x, G[x].push_back(cnt);
        while (1) {
          int u = st.top();
          st.pop();
          p[u] = cnt, G[cnt].push_back(u);
          if (u == g[x][i]) break;
        }
      }
    }
  }
}
int cnt1, h[200004], siz[200004], cmp[200004], nxt[200004];
void dfs(int x) {
  siz[x] = 1;
  int maxx = 0;
  for (int i = 0; i < G[x].size(); i++) {
    dfs(G[x][i]), siz[x] += siz[G[x][i]];
    if (maxx < siz[G[x][i]]) h[x] = G[x][i], maxx = siz[G[x][i]];
  }
}
void dfs1(int x, int H) {
  cmp[x] = ++cnt1, nxt[x] = H;
  if (h[x]) dfs1(h[x], H);
  for (int i = 0; i < G[x].size(); i++)
    if (G[x][i] != h[x]) dfs1(G[x][i], G[x][i]);
}
int cmp1[200004], L[200004], R[200004], cnt2;
void bfs() {
  queue<int> que;
  que.push(1), cmp1[1] = 1, cnt2 = 1;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    L[u] = cnt2 + 1;
    for (int i = 0; i < G[u].size(); i++)
      cmp1[G[u][i]] = ++cnt2, que.push(G[u][i]);
    R[u] = cnt2;
  }
}
int cnt_ = 1, dat[525000], dat1[525000];
void upd(int *d, int id, int x) {
  d[id] = x;
  while (id) {
    id = (id - 1) / 2;
    d[id] = min(d[id * 2 + 1], d[id * 2 + 2]);
  }
}
int got(int *d, int l, int r, int id, int a, int b) {
  if (r < a || b < l) return 1e9;
  if (a <= l && r <= b) return d[id];
  return min(got(d, l, (l + r) / 2, id * 2 + 1, a, b),
             got(d, (l + r) / 2 + 1, r, id * 2 + 2, a, b));
}
void built(int *d, int l, int r, int id) {
  if (l > cnt) return;
  if (l == r) return;
  built(d, l, (l + r) / 2, id * 2 + 1),
      built(d, (l + r) / 2 + 1, r, id * 2 + 2);
  d[id] = min(d[id * 2 + 1], d[id * 2 + 2]);
}
char opt[3];
bool isans(int l, int r, int x) { return cmp[l] <= cmp[x] && cmp[x] <= cmp[r]; }
int fans(int u, int v) {
  int ret = 1e9;
  while (1) {
    if (cmp[u] > cmp[v]) swap(u, v);
    int L = nxt[v];
    if (isans(L, v, u)) {
      ret = min(ret, got(dat1, 1, cnt_, 0, cmp[u], cmp[v]));
      if (u > n) ret = min(w[p[u]], ret);
      return ret;
    }
    ret = min(ret, got(dat1, 1, cnt_, 0, cmp[L], cmp[v]));
    v = p[L];
  }
}
int main() {
  scanf("%d%d%d", &n, &m, &Q);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v), g[v].push_back(u);
  }
  cnt = n;
  tarjan(1);
  dfs(1);
  dfs1(1, 1);
  bfs();
  while (cnt_ < cnt) cnt_ <<= 1;
  for (int i = 1; i <= n; i++)
    dat[cmp1[i] + cnt_ - 2] = dat1[cmp[i] + cnt_ - 2] = w[i];
  built(dat, 1, cnt_, 0);
  for (int i = n + 1; i <= cnt; i++) {
    int ret = got(dat, 1, cnt_, 0, L[i], R[i]);
    dat1[cmp[i] + cnt_ - 2] = ret;
  }
  built(dat1, 1, cnt_, 0);
  while (Q--) {
    int u, v;
    scanf("%s%d%d", opt, &u, &v);
    if (*opt == 'C') {
      w[u] = v;
      upd(dat, cmp1[u] + cnt_ - 2, v);
      upd(dat1, cmp[u] + cnt_ - 2, v);
      if (u > 1) {
        int par = p[u];
        int ret = got(dat, 1, cnt_, 0, L[par], R[par]);
        upd(dat1, cmp[par] + cnt_ - 2, ret);
      }
    } else
      write(fans(u, v)), puts("");
  }
}