C. Rings---Codeforces Round #741 (Div. 2)

C. Rings

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard outputFrodo was caught by Saruman. He tore a pouch from Frodo's neck, shook out its contents —there was a pile of different rings: gold and silver...

"How am I to tell which is the One?!" the mage howled.

"Throw them one by one into the Cracks of Doom and watch when Mordor falls!"

Somewhere in a parallel Middle-earth, when Saruman caught Frodo, he only found nn rings. And the ii-th ring was either gold or silver. For convenience Saruman wrote down a binary string ss of nn characters, where the ii-th character was 0 if the ii-th ring was gold, and 1 if it was silver.

Saruman has a magic function ff, which takes a binary string and returns a number obtained by converting the string into a binary number and then converting the binary number into a decimal number. For example, f(001010)=10,f(111)=7,f(11011101)=221f(001010)=10,f(111)=7,f(11011101)=221.

Saruman, however, thinks that the order of the rings plays some important role. He wants to find 22 pairs of integers (l1,r1),(l2,r2)(l1,r1),(l2,r2), such that:

• 1≤l1≤n1≤l1≤n, 1≤r1≤n1≤r1≤n, r1−l1+1≥⌊n2⌋r1−l1+1≥⌊n2⌋
• 1≤l2≤n1≤l2≤n, 1≤r2≤n1≤r2≤n, r2−l2+1≥⌊n2⌋r2−l2+1≥⌊n2⌋
• Pairs (l1,r1)(l1,r1) and (l2,r2)(l2,r2) are distinct. That is, at least one of l1≠l2l1≠l2 and r1≠r2r1≠r2 must hold.
• Let tt be the substring s[l1:r1]s[l1:r1] of ss, and ww be the substring s[l2:r2]s[l2:r2] of ss. Then there exists non-negative integer kk, such that f(t)=f(w)⋅kf(t)=f(w)⋅k.

Here substring s[l:r]s[l:r] denotes slsl+1…sr−1srslsl+1…sr−1sr, and ⌊x⌋⌊x⌋ denotes rounding the number down to the nearest integer.

Help Saruman solve this problem! It is guaranteed that under the constraints of the problem at least one solution exists.

Input

Each test contains multiple test cases.

The first line contains one positive integer tt (1≤t≤1031≤t≤103), denoting the number of test cases. Description of the test cases follows.

The first line of each test case contains one positive integer nn (2≤n≤2⋅1042≤n≤2⋅104) — length of the string.

The second line of each test case contains a non-empty binary string of length nn.

It is guaranteed that the sum of nn over all test cases does not exceed 105105.

Output

For every test case print four integers l1l1, r1r1, l2l2, r2r2, which denote the beginning of the first substring, the end of the first substring, the beginning of the second substring, and the end of the second substring, respectively.

If there are multiple solutions, print any.

Example

input

Copy

7
6
101111
9
111000111
8
10000000
5
11011
6
001111
3
101
30
100000000000000100000000000000

output

Copy

3 6 1 3
1 9 4 9
5 8 1 4
1 5 3 5
1 6 2 4
1 2 2 3
1 15 16 30

Note

In the first testcase f(t)=f(1111)=15f(t)=f(1111)=15, f(w)=f(101)=5f(w)=f(101)=5.

In the second testcase f(t)=f(111000111)=455f(t)=f(111000111)=455, f(w)=f(000111)=7f(w)=f(000111)=7.

In the third testcase f(t)=f(0000)=0f(t)=f(0000)=0, f(w)=f(1000)=8f(w)=f(1000)=8.

In the fourth testcase f(t)=f(11011)=27f(t)=f(11011)=27, f(w)=f(011)=3f(w)=f(011)=3.

In the fifth testcase f(t)=f(001111)=15f(t)=f(001111)=15, f(w)=f(011)=3f(w)=f(011)=3.

=========================================================================

首先全是1肯定是满足的，选择1-n-1 2-n大小相等

然后我们注意到0的位置可能在1--n/2

n/2+1 --n 

一旦在1--n/2,那么该位置 p--n  p+1--n是一定相等且长度都满足要求的

如果在n/2+1 --- n  那么 1-p   1-p-1是二倍的关系 （二进制后面添加0会扩大二倍，也称为整体左移一位，是 <<运算 )

所以一定是能够满足的

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
 
 
int main()
{
 
    int t;
 
    cin>>t;
 
    while(t--)
    {
        int n;
 
        cin>>n;
 
        string s;
 
        cin>>s;
        // 0 1 2 3 4
 
        //0 1 2 3
 
        //0 1 2
 
        //0 1
 
        int flag=0;
 
        s=" "+s;
 
        for(int i=1;i<=n;i++)
        {
            if(s[i]=='0')
            {
 
                if(i>n/2)
                {
                    cout<<1<<" "<" "<<1<<" "<-1<
 
                    flag=1;
 
                    break;
                }
                else
                {
                    cout<" "<" "<1<<" "<
 
                    flag=1;
 
                    break;
                }
            }
        }
 
        if(!flag)
        {
            cout<<1<<" "<-1<<" "<<2<<" "<
        }
    }
    return 0;
}