Although Inzane successfully found his beloved bone, Zane, his owner, has yet to return. To search for Zane, he would need a lot of money, of which he sadly has none. To deal with the problem, he has decided to hack the banks.
There are n banks, numbered from 1 to n. There are also n - 1 wires connecting the banks. All banks are initially online. Each bank also has its initial strength: bank i has initial strength ai.
Let us define some keywords before we proceed. Bank i and bank j are neighboring if and only if there exists a wire directly connecting them. Bank i and bank j are semi-neighboring if and only if there exists an online bank k such that bank i and bank k are neighboring and bank k and bank j are neighboring.
When a bank is hacked, it becomes offline (and no longer online), and other banks that are neighboring or semi-neighboring to it have their strengths increased by 1.
To start his plan, Inzane will choose a bank to hack first. Indeed, the strength of such bank must not exceed the strength of his computer. After this, he will repeatedly choose some bank to hack next until all the banks are hacked, but he can continue to hack bank x if and only if all these conditions are met:
- Bank x is online. That is, bank x is not hacked yet.
- Bank x is neighboring to some offline bank.
- The strength of bank x is less than or equal to the strength of Inzane's computer.
Determine the minimum strength of the computer Inzane needs to hack all the banks.
The first line contains one integer n (1 ≤ n ≤ 3·105) — the total number of banks.
The second line contains n integers a1, a2, ..., an ( - 109 ≤ ai ≤ 109) — the strengths of the banks.
Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — meaning that there is a wire directly connecting banks ui and vi.
It is guaranteed that the wires connect the banks in such a way that Inzane can somehow hack all the banks using a computer with appropriate strength.
Print one integer — the minimum strength of the computer Inzane needs to accomplish the goal.
5
1 2 3 4 5
1 2
2 3
3 4
4 5
5
7
38 -29 87 93 39 28 -55
1 2
2 5
3 2
2 4
1 7
7 6
93
5
1 2 7 6 7
1 5
5 3
3 4
2 4
8
In the first sample, Inzane can hack all banks using a computer with strength 5. Here is how:
- Initially, strengths of the banks are [1, 2, 3, 4, 5].
- He hacks bank 5, then strengths of the banks become [1, 2, 4, 5, - ].
- He hacks bank 4, then strengths of the banks become [1, 3, 5, - , - ].
- He hacks bank 3, then strengths of the banks become [2, 4, - , - , - ].
- He hacks bank 2, then strengths of the banks become [3, - , - , - , - ].
- He completes his goal by hacking bank 1.
In the second sample, Inzane can hack banks 4, 2, 3, 1, 5, 7, and 6, in this order. This way, he can hack all banks using a computer with strength 93.
第一次可以随便选择一个攻击，其余的必须遵守规定来攻击；
考虑如下情况：
如果最大值 maxx只有一个，且值为 maxx-1 的数量=与maxx相邻的maxx-1的数量，那么耗费的 hack值= maxx，否则应该为 maxx+1;
如果最大值 maxx不止一个，且存在一个点使得该点相邻的 maxx的数量=maxx的数量，那么hack=maxx+1，否则为 maxx+2;
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-4
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
int n;
int a[maxn];
vector<int>vc[maxn];
int main() {
//ios::sync_with_stdio(0);
rdint(n);
for (int i = 1; i <= n; i++)rdint(a[i]);
for (int i = 1; i < n; i++) {
int u, v; rdint(u); rdint(v);
vc[u].push_back(v); vc[v].push_back(u);
}
int maxx = -inf; int maxcnt = 0;
for (int i = 1; i <= n; i++) {
maxx = max(maxx, a[i]);
}
for (int i = 1; i <= n; i++)if (a[i] == maxx)maxcnt++;
if (maxcnt == 1) {
int mxcnt = 0;
int pos = 0;
for (int i = 1; i <= n; i++) {
if (a[i] == maxx - 1)mxcnt++;
if (a[i] == maxx)pos = i;
}
int num = 0;
for (int i = 0; i < vc[pos].size(); i++) {
if (a[vc[pos][i]] == maxx - 1)num++;
}
if (num == mxcnt) { cout << maxx << endl; return 0; }
else {
cout << maxx + 1 << endl; return 0;
}
}
else {
bool fg = 0;
for (int i = 1; i <= n; i++) {
int num = 0;
if (a[i] == maxx)num++;
for (int j = 0; j < vc[i].size(); j++) {
if (a[vc[i][j]] == maxx)num++;
}
if (num == maxcnt)fg = 1;
}
if (fg)cout << maxx + 1 << endl;
else cout << maxx + 2 << endl;
}
return 0;
}