In this CodeChef copy and paste, problem solution Chef has binary string A of length N. He constructs a new binary string B by concatenating M copies of A together. For example, if A="10010", M = 3, then B="100101001010010"

CODECHEF Copy and Paste Problem Solution (NCOPIES)


CODECHEF Copy and Paste Problem Solution (NCOPIES)

//Divyansh Vinayak
#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",x)
#define pl(x)	printf("%lld",x)
#define ps(s)	printf("%s",s)
#define pnl()   printf("\n")
#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;
int mpow(int base, int exp); 
void ipgraph(int m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 18 * 103, M = N;
#define top(a,b) (ll)((a+b-1)/b)
//=======================
vi g[N];
void matrix(ll n, ll m){
  char s[101][101];
    while(cin>>n>>m){
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cin>>s[i][j];
            }
        }
}
}

int fact(ll n);
 

 
// Returns factorial of n
int fact(ll n)
{
    ll res = 1;
    for (int i = 2; i <= n; i++)
        res = res * i;
    return res;
}

int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}

int main() {
  ll t;
  cin >> t;
  while(t--){
  ll x = 0,y = 0,z = 0;
  char s[2000002];
  ll n , m;
  cin >> n >> m >> (s+1);
  for(int i=1;i<=n;++i)
    if(s[i]=='1') x++;
  if(x==0) {
    cout << n*m << endl;
    continue;
  }
  if(x*m%2==1) {
    cout << 0 << endl;
    continue;
  }
  if(m%2==0) {
    for(int i=1;i<=n;++i) {
      if(s[i]=='1') break;
      y++;
    }
    for(int i=n;i>=1;--i) {
      if(s[i]=='1') break;
      y++;
    }
    cout << y+1 << endl;
    continue;
  }
  for(int i=1;i<=n;++i) {
    if(s[i]=='1') z++;
    if(z==x/2) y++;
  }
  cout << y << endl;

  }
  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;
		g[u-1].pb(v-1);
		g[v-1].pb(u-1);
	}
}

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