D. Grid-00100

------------恢复内容开始------------

D. Grid-00100

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A mad scientist Dr.Jubal has made a competitive programming task. Try to solve it!

You are given integers n,kn,k. Construct a grid AA with size n×nn×n consisting of integers 00 and 11. The very important condition should be satisfied: the sum of all elements in the grid is exactly kk. In other words, the number of 11 in the grid is equal to kk.

Let's define:

• Ai,jAi,j as the integer in the ii-th row and the jj-th column.
• Ri=Ai,1+Ai,2+...+Ai,nRi=Ai,1+Ai,2+...+Ai,n (for all 1≤i≤n1≤i≤n).
• Cj=A1,j+A2,j+...+An,jCj=A1,j+A2,j+...+An,j (for all 1≤j≤n1≤j≤n).
• In other words, RiRi are row sums and CjCj are column sums of the grid AA.
• For the grid AA let's define the value f(A)=(max(R)−min(R))2+(max(C)−min(C))2f(A)=(max(R)−min(R))2+(max(C)−min(C))2 (here for an integer sequence XX we define max(X)max(X) as the maximum value in XX and min(X)min(X) as the minimum value in XX).

Find any grid AA, which satisfies the following condition. Among such grids find any, for which the value f(A)f(A) is the minimum possible. Among such tables, you can find any.

Input

The input consists of multiple test cases. The first line contains a single integer tt (1≤t≤1001≤t≤100) — the number of test cases. Next tt lines contain descriptions of test cases.

For each test case the only line contains two integers nn, kk (1≤n≤300,0≤k≤n2)(1≤n≤300,0≤k≤n2).

It is guaranteed that the sum of n2n2 for all test cases does not exceed 105105.

Output

For each test case, firstly print the minimum possible value of f(A)f(A) among all tables, for which the condition is satisfied.

After that, print nn lines contain nn characters each. The jj-th character in the ii-th line should be equal to Ai,jAi,j.

If there are multiple answers you can print any.

Example

input

Copy

4
2 2
3 8
1 0
4 16

output

Copy

0
10
01
2
111
111
101
0
0
0
1111
1111
1111
1111

Note

In the first test case, the sum of all elements in the grid is equal to 22, so the condition is satisfied. R1=1,R2=1R1=1,R2=1 and C1=1,C2=1C1=1,C2=1. Then, f(A)=(1−1)2+(1−1)2=0f(A)=(1−1)2+(1−1)2=0, which is the minimum possible value of f(A)f(A).

In the second test case, the sum of all elements in the grid is equal to 88, so the condition is satisfied. R1=3,R2=3,R3=2R1=3,R2=3,R3=2 and C1=3,C2=2,C3=3C1=3,C2=2,C3=3. Then, f(A)=(3−2)2+(3−2)2=2f(A)=(3−2)2+(3−2)2=2. It can be proven, that it is the minimum possible value of f(A)f(A).

------------恢复内容结束------------

#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <queue>
#include <map>
#include <sstream>
#include <cstdio>
#include <cstring>
#include <numeric>
#include <cmath>
#include <iomanip>
#include <deque>
#include <bitset>
//#include <unordered_set>
//#include <unordered_map>
#define ll              long long
#define pii             pair<int, int>
#define rep(i,a,b)      for(int  i=a;i<=b;i++)
#define dec(i,a,b)      for(int  i=a;i>=b;i--)
#define forn(i, n)      for(int i = 0; i < int(n); i++)
using namespace std;
int dir[4][2] = { { 1,0 },{ 0,1 } ,{ 0,-1 },{ -1,0 } };
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846;
const double eps = 1e-6;
const int mod = 1e9 + 7;
const int N = 1e5 + 5;
//if(x<0 || x>=r || y<0 || y>=c)

inline ll read()
{
    ll x = 0; bool f = true; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
    while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
    return f ? x : -x;
}
ll gcd(ll m, ll n)
{
    return n == 0 ? m : gcd(n, m % n);
}
ll lcm(ll m, ll n)
{
    return m * n / gcd(m, n);
}
bool prime(int x) {
    if (x < 2) return false;
    for (int i = 2; i * i <= x; ++i) {
        if (x % i == 0) return false;
    }
    return true;
}
ll qpow(ll m, ll k, ll mod)
{
    ll res = 1, t = m;
    while (k)
    {
        if (k & 1)
            res = res * t % mod;
        t = t * t % mod;
        k >>= 1;
    }
    return res;
}

int main()
{
    int T;
    cin >> T;
    while (T--)
    {
        int n, k, t, i, j, p, q;
        char res[305][305];
        cin >> n >> k;
        if (k % n == 0) { printf("0
"); }
        else { printf("2
"); }
        for (i = 0; i < n; i++) {
            for (j = 0; j < n; j++) {
                res[i][j] = '0';
            }
            res[i][n] = 0;
        }
        p = 0; q = 0;
        while (k > 0) {
            k--;
            res[p][q] = '1';
            p++; q++; q %= n;
            if (p == n) {
                p = 0; q++; q %= n;
            }
        }
        for (i = 0; i < n; i++) {
            printf("%s
", res[i]);
        }
    }
    return 0;
}