You are given a string s, consisting of n letters, each letter is either 'a' or 'b'. The letters in the string are numbered from 1 to n.

s[l; r] is a continuous substring of letters from index l to r of the string inclusive. 

A string is called balanced if the number of letters 'a' in it is equal to the number of letters 'b'. For example, strings "baba" and "aabbab" are balanced and strings "aaab" and "b" are not.

Find any non-empty balanced substring s[l; r] of string s. Print its l and r (1 ≤ l ≤ r ≤ n). If there is no such substring, then print -1 -1.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.

Then the descriptions of t testcases follow.

The first line of the testcase contains a single integer n (1 ≤ n ≤ 50) — the length of the string.

The second line of the testcase contains a string s, consisting of n letters, each letter is either 'a' or 'b'.

Output

For each testcase print two integers. If there exists a non-empty balanced substring s[l; r], then print l r (1 ≤ l ≤ r ≤ n). Otherwise, print -1 -1.

Example

Input


4
1
a
6
abbaba
6
abbaba
9
babbabbaa


Output


-1 -1
1 6
3 6
2 5

Note

In the first testcase there are no non-empty balanced subtrings.

In the second and third testcases there are multiple balanced substrings, including the entire string "abbaba" and substring "baba".
/* || गुरु गृह गए पढ़न रघुराई अल्प काल विद्या सब आई || */
/* || Guru grihe gaye padhan Raghurai, alpkaal vidhya sab aayi || */

#include <bits/stdc++.h>
using namespace std;
#define gc getchar_unlocked
#define fo(i, n) for (i = 0; i < n; i++)
#define Fo(i, k, n) for (i = k; k < n ? i < n : i > n; k < n ? i += 1 : i -= 1)
#define ll long long
#define si(x) scanf("%d", &x)
#define sl(x) scanf("%lld", &x)
#define ss(s) scanf("%s", s)
#define pi(x) printf("%d\n", x)
#define pl(x) printf("%lld\n", x)
#define ps(s) printf("%s\n", s)
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for (auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
typedef pair<int, int> pii;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pii> vpii;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim)
{
    uniform_int_distribution<int> uid(0, lim - 1);
    return uid(rang);
}
int mpow(int base, int exp);
void ipgraph(int n, int m);
void dfs(int u, int par);

const int mod = 1'000'000'007;
const int N = 3e5, M = N;
//=======================

vi g[N];
int a[N];

void solve()
{
    int i, j, n, m;
    cin >> n;
    string s;
    cin >> s;

    fo(i, n - 1)
    {
        int ctr = 0;
        if (s[i] == 'a')
            ctr++;
        else
            ctr--;
        for (j = i + 1; j < n; ++j)
        {
            if (s[j] == 'a')
                ctr++;
            else
                ctr--;

            if (ctr == 0)
            {
                cout << i << " " << j << endl;
                return;
            }
        }
    }

    cout << -1 << " " << -1 << endl;
}

int main()
{
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    srand(chrono::high_resolution_clock::now().time_since_epoch().count());

    int t = 1;
    cin >> t;
    while (t--)
    {
        solve();
    }

    return 0;
}

int mpow(int base, int exp)
{
    base %= mod;
    int result = 1;
    while (exp > 0)
    {
        if (exp & 1)
            result = ((ll)result * base) % mod;
        base = ((ll)base * base) % mod;
        exp >>= 1;
    }
    return result;
}

void ipgraph(int n, int m)
{
    int i, u, v;
    while (m--)
    {
        cin >> u >> v;
        u--, v--;
        g[u].pb(v);
        g[v].pb(u);
    }
}

void dfs(int u, int par)
{
    for (int v : g[u])
    {
        if (v == par)
            continue;
        dfs(v, u);
    }
}

bool getBit(int num, int i)
{
    return ((num & (1 << i)) != 0);
}

int setBit(int num, int i)
{
    return num | (1 << i);
}

int clearBit(int num, int i)
{
    int mask = ~(1 << i);
    return num & mask;
}