Problem: This is an interactive problem. Remember to flush your output while communicating with the testing program. You may use fflush(stdout) in C++, system.out.flush() in Java, stdout.flush() in Python or flush(output) in Pascal to flush the output. If you use some other programming language, consult its documentation. You may also refer to the guide on interactive problems: <https://codeforces.com/blog/entry/45307>.

There is a city in which Dixit lives. In the city, there are n houses. There is  exactly one directed road between every pair of houses. For example, consider two houses A and B, then there is a directed road either from A to B or from B to A but not both. The number of roads leading to the i-th house is k_i.

Two houses A and B are bi-reachable if A is reachable from B and B is reachable from A. We say that house B is reachable from house A when there is a path from house A to house B.

Dixit wants to buy two houses in the city, that is, one for living and one for studying. Of course, he would like to travel from one house to another. So, he wants to find a pair of bi-reachable houses A and B. Among all such pairs, he wants to choose one with the maximum value of |k_A - k_B|, where k_i is the number of roads leading to the house i. If more than one optimal pair exists, any of them is suitable.

Since Dixit is busy preparing CodeCraft, can you help him find the desired pair of houses, or tell him that no such houses exist?

In the problem input, you are not given the direction of each road. You are given — for each house — only the number of incoming roads to that house (k_i).

You are allowed to ask only one type of query from the judge: give two houses A and B, and the judge answers whether B is reachable from A. There is no upper limit on the number of queries. But, you cannot ask more queries after the judge answers "Yes" to any of your queries. Also, you cannot ask the same query twice.

Once you have exhausted all your queries (or the judge responds "Yes" to any of your queries), your program must output its guess for the two houses and quit.

See the Interaction section below for more details.

Input

The first line contains a single integer n (3 ≤ n ≤ 500) denoting the number of houses in the city. The next line contains n space-separated integers k_1, k_2, ..., k_n (0 ≤ k_i ≤ n - 1), the i-th of them represents the number of incoming roads to the i-th house.

Interaction

To ask a query, print "? A B" (1 ≤ A,B ≤ N, A≠ B). The judge will respond "Yes" if house B is reachable from house A, or "No" otherwise.

To output the final answer, print "! A B", where A and B are bi-reachable with the maximum possible value of |k_A - k_B|. If there does not exist such pair of houses A and B, output "! 0 0".

After outputting the final answer, your program must terminate immediately, otherwise you will receive Wrong Answer verdict.

You cannot ask the same query twice. There is no upper limit to the number of queries you ask, but, you cannot ask more queries after the judge answers "Yes" to any of your queries. Your program must now output the final answer ("! A B" or "! 0 0") and terminate.

If you ask a query in incorrect format or repeat a previous query, you will get Wrong Answer verdict.

After printing a query do not forget to output the end of the line and flush the output. Otherwise, you will get the Idleness limit exceeded error. To do this, use:

  * fflush(stdout) or cout.flush() in C++; 
  * System.out.flush() in Java; 
  * flush(output) in Pascal; 
  * stdout.flush() in Python; 
  * see documentation for other languages.

Examples

Input


3
1 1 1
Yes

Output


? 1 2
! 1 2

Input


4
1 2 0 3
No
No
No
No
No
No

Output


? 2 1
? 1 3
? 4 1
? 2 3
? 4 2
? 4 3
! 0 0

Note

In the first sample input, we are given a city of three houses with one incoming road each. The user program asks one query: "? 1 2": asking whether we can reach from house 1 to house 2. The judge responds with "Yes". The user program now concludes that this is sufficient information to determine the correct answer. So, it outputs "! 1 2" and quits.

In the second sample input, the user program queries for six different pairs of houses, and finally answers "! 0 0" as it is convinced that no two houses as desired in the question exist in this city.

Solution (in CPP):

Solution:
/*
 * Kon Kon Fubuki
*/

#include <bits/stdc++.h>

using namespace std;

//basic types
using ll = long long;
using ld = long double;
using db = double;
using str = string;

using pi = pair<int,int>;
using pl = pair<ll,ll>;
using pd = pair<db,db>;

using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;

#define tcT template<class T
#define tcTU tcT, class U

tcT> using V = vector<T>; 
tcT, size_t SZ> using AR = array<T,SZ>; 
tcT> using PR = pair<T,T>;

// pairs
#define mp make_pair
#define f first
#define s second

// for vectors
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend()
#define sor(x) sort(all(x))
#define rsz resize
#define ins insert
#define ft front()
#define bk back()
#define pf push_front
#define pb push_back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound

// loops
#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); i++)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7;
const int mxN = 5e2+5;
const ll INF = 1e18;
const ld PI = acos((ld)-1);
const int tSZ = (1 << 21);
const int dx[4] = {1,0,-1,0}, dy[4] = {0,1,0,-1}; // for grid problems
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 

// useful functions
constexpr int pcnt(int x) { return __builtin_popcount(x); } // number of bits
constexpr int bits(int x) { return 31-__builtin_clz(x); } // log2(x) rounded down
ll cdiv(ll a, ll b) { return a/b+((a^b)>0&&a%b); } // divide a by b rounded up
ll fdiv(ll a, ll b) { return a/b-((a^b)<0&&a%b); } // divide a by b rounded down

// functions for working with MOD
ll add(ll f, ll s){ f += s; if(f >= MOD) f %= MOD; if(f < 0) f += MOD;	return f; } // add f to s and make sure it's in range
ll mul(ll f, ll s){ f *= s; if(f >= MOD) f %= MOD; return f;} // multiply f and s and make sure it's in range
ll pw(ll x, ll y) { ll ans = 1, ml = x; while(y) { if(y & 1) ans = mul(ans, ml); ml = mul(ml, ml);	y >>= 1; } return ans; } // number x to a power y
ll inv(ll x) { return pw(x, MOD - 2); } //inversion of a number

tcT> void remDup(vector<T>& x) { // sort and remove duplicates
	sort(all(x)); x.erase(unique(all(x)),end(x)); }
tcTU> void erase(T& x, const U& v) {
	auto it = x.find(v); assert(it != end(x));
	x.erase(v); }

// Input
#define tcTUU tcT, class ...U
tcT> void re(complex<T>& c);
tcTU> void re(pair<T,U>& p);
tcT> void re(vector<T>& v);
tcT, size_t SZ> void re(array<T,SZ>& a);

tcT> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }
 
tcT> void re(complex<T>& c) { T a,b; re(a,b); c = {a,b}; }
tcTU> void re(pair<T,U>& p) { re(p.f,p.s); }
tcT> void re(vector<T>& x) { trav(a,x) re(a); }
tcT, size_t SZ> void re(array<T,SZ>& x) { trav(a,x) re(a); }
tcT> void rv(int& n, vector<T>& x) { re(n); x.rsz(n); trav(a,x) re(a); }

// To_String
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(bool b) { 
	#ifdef LOCAL
		return b ? "true" : "false"; 
	#else 
		return ts((int)b);
	#endif
}
tcT> str ts(complex<T> c) { 
	stringstream ss; ss << c; return ss.str(); }
str ts(vector<bool> v) {
	str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]);
	res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
tcTU> str ts(pair<T,U> p);
tcT> str ts(T v) { // containers with begin(), end()
	#ifdef LOCAL
		bool fst = 1; str res = "{";
		for (const auto& x: v) {
			if (!fst) res += ", ";
			fst = 0; res += ts(x);
		}
		res += "}"; return res;
	#else
		bool fst = 1; str res = "";
		for (const auto& x: v) {
			if (!fst) res += " ";
			fst = 0; res += ts(x);
		}
		return res;
 
	#endif
}
tcTU> str ts(pair<T,U> p) {
	#ifdef LOCAL
		return "("+ts(p.f)+", "+ts(p.s)+")"; 
	#else
		return ts(p.f)+" "+ts(p.s);
	#endif
}
 
// Output
tcT> void pr(T v) { cout << ts(v); }
tcTUU> void pr(const T& t, const U&... u) { 
	pr(t); pr(u...); }
void ps() { pr("\n"); } // no spaces
tcTUU> void ps(const T& t, const U&... u) { 
	pr(t); if (sizeof...(u)) pr(" "); ps(u...); }
 
// File In/Out
void setIn(str s) { freopen(s.c_str(),"r", stdin); }
void setOut(str s) { freopen(s.c_str(),"w", stdout); }
void unsyncIO() { cin.tie(0)->sync_with_stdio(0); }
void setIO(str st = "") {
	unsyncIO();
	if (sz(st)) { setIn(st+".in"), setOut(st+".out"); } // If problem requires file reading
}

int n, a[mxN];

int ask(int x, int y)
{
	ps("?", x, y);
	cout.flush();
	str ans; re(ans);
	return ans[0] == 'Y';
}

int aa[mxN][mxN];
vi grph[mxN];
int cc[mxN][mxN];

void solve()
{
	priority_queue <array<int, 3>> pq;
	FOR(i, 1, n + 1)
	{
		FOR(j, i + 1, n + 1)
		{
			if(a[i] > a[j])
			{
				pq.push({a[i] - a[j], i, j});
			}
			else
			{
				pq.push({a[j] - a[i], j, i});
			}
		}
	}
	while(!pq.empty())
	{
		int ff = pq.top()[1];
		int ss = pq.top()[2];
		pq.pop();
		if(ask(ff, ss))
		{
			ps("!", ff, ss);
			return;
		}
	}
	ps("!", 0, 0);
}

int main()
{
	setIO();

	re(n);
	FOR(i, 1, n + 1)
	{
		re(a[i]);
	}
	solve();

	return 0;
	//read stuff at the bottom ffs
}
/* things to keep in mind 
 * int overflow, array bounds
 * any special cases
 * always do something
 * WRITE STUFF DOWN
 * THINK ABOUT OTHER APPROACHES
 * DON'T NON STOP CHECK OTHERS
 * DON'T PANIC
*/