# CODECHEF Copy and Paste Problem Solution (NCOPIES)

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)

```//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);
}
}
```