Problem: The tycoon of a winery empire in Mondstadt, unmatched in every possible way. A thinker in the Knights of Favonius with an exotic appearance.

This time, the brothers are dealing with a strange piece of wood marked with their names. This plank of wood can be represented as a string of n characters. Each character is either a 'D' or a 'K'. You want to make some number of cuts (possibly 0) on this string, partitioning it into several contiguous pieces, each with length at least 1. Both brothers act with dignity, so they want to split the wood as evenly as possible. They want to know the maximum number of pieces you can split the wood into such that the ratios of the number of occurrences of 'D' to the number of occurrences of 'K' in each chunk are the same.

Kaeya, the curious thinker, is interested in the solution for multiple scenarios. He wants to know the answer for every prefix of the given string. Help him to solve this problem!

For a string we define a ratio as a:b where 'D' appears in it a times, and 'K' appears b times. Note that a or b can equal 0, but not both. Ratios a:b and c:d are considered equal if and only if a⋅ d = b⋅ c. 

For example, for the string 'DDD' the ratio will be 3:0, for 'DKD' — 2:1, for 'DKK' — 1:2, and for 'KKKKDD' — 2:4. Note that the ratios of the latter two strings are equal to each other, but they are not equal to the ratios of the first two strings.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 1000). Description of the test cases follows.

The first line of each test case contains an integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the length of the wood.

The second line of each test case contains a string s of length n. Every character of s will be either 'D' or 'K'.

It is guaranteed that the sum of n over all test cases does not exceed 5 ⋅ 10^5.

Output

For each test case, output n space separated integers. The i-th of these numbers should equal the answer for the prefix s_{1},s_{2},...,s_{i}.

Example

Input


5
3
DDK
6
DDDDDD
4
DKDK
1
D
9
DKDKDDDDK


Output


1 2 1 
1 2 3 4 5 6 
1 1 1 2 
1 
1 1 1 2 1 2 1 1 3 

Note

For the first test case, there is no way to partition 'D' or 'DDK' into more than one block with equal ratios of numbers of 'D' and 'K', while you can split 'DD' into 'D' and 'D'.

For the second test case, you can split each prefix of length i into i blocks 'D'.

Solution (in CPP):

Solution:
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("inline")
#include<bits/stdc++.h>
using namespace std;
template<class T> struct cLtraits_identity{
  using type = T;
}
;
template<class T> using cLtraits_try_make_signed =
  typename conditional<
    is_integral<T>::value,
    make_signed<T>,
    cLtraits_identity<T>
    >::type;
struct Rand{
  unsigned x;
  unsigned y;
  unsigned z;
  unsigned w;
  Rand(void){
    x=123456789;
    y=362436069;
    z=521288629;
    w=(unsigned)time(NULL);
  }
  Rand(unsigned seed){
    x=123456789;
    y=362436069;
    z=521288629;
    w=seed;
  }
  inline unsigned get(void){
    unsigned t;
    t = (x^(x<<11));
    x=y;
    y=z;
    z=w;
    w = (w^(w>>19))^(t^(t>>8));
    return w;
  }
  inline double getUni(void){
    return get()/4294967296.0;
  }
  inline int get(int a){
    return (int)(a*getUni());
  }
  inline int get(int a, int b){
    return a+(int)((b-a+1)*getUni());
  }
  inline long long get(long long a){
    return(long long)(a*getUni());
  }
  inline long long get(long long a, long long b){
    return a+(long long)((b-a+1)*getUni());
  }
  inline double get(double a, double b){
    return a+(b-a)*getUni();
  }
  inline int getExp(int a){
    return(int)(exp(getUni()*log(a+1.0))-1.0);
  }
  inline int getExp(int a, int b){
    return a+(int)(exp(getUni()*log((b-a+1)+1.0))-1.0);
  }
}
;
inline int my_getchar(){
  static char buf[1048576];
  static int s = 1048576;
  static int e = 1048576;
  if(s == e && e == 1048576){
    e = fread(buf, 1, 1048576, stdin);
    s = 0;
  }
  if(s == e){
    return EOF;
  }
  return buf[s++];
}
inline void rd(int &x){
  int k;
  int m=0;
  x=0;
  for(;;){
    k = my_getchar();
    if(k=='-'){
      m=1;
      break;
    }
    if('0'<=k&&k<='9'){
      x=k-'0';
      break;
    }
  }
  for(;;){
    k = my_getchar();
    if(k<'0'||k>'9'){
      break;
    }
    x=x*10+k-'0';
  }
  if(m){
    x=-x;
  }
}
inline void rd(char &c){
  int i;
  for(;;){
    i = my_getchar();
    if(i!=' '&&i!='\n'&&i!='\r'&&i!='\t'&&i!=EOF){
      break;
    }
  }
  c = i;
}
inline int rd(char c[]){
  int i;
  int sz = 0;
  for(;;){
    i = my_getchar();
    if(i!=' '&&i!='\n'&&i!='\r'&&i!='\t'&&i!=EOF){
      break;
    }
  }
  c[sz++] = i;
  for(;;){
    i = my_getchar();
    if(i==' '||i=='\n'||i=='\r'||i=='\t'||i==EOF){
      break;
    }
    c[sz++] = i;
  }
  c[sz]='\0';
  return sz;
}
inline int rd_int(void){
  int x;
  rd(x);
  return x;
}
struct MY_WRITER{
  char buf[1048576];
  int s;
  int e;
  MY_WRITER(){
    s = 0;
    e = 1048576;
  }
  ~MY_WRITER(){
    if(s){
      fwrite(buf, 1, s, stdout);
    }
  }
}
;
MY_WRITER MY_WRITER_VAR;
void my_putchar(int a){
  if(MY_WRITER_VAR.s == MY_WRITER_VAR.e){
    fwrite(MY_WRITER_VAR.buf, 1, MY_WRITER_VAR.s, stdout);
    MY_WRITER_VAR.s = 0;
  }
  MY_WRITER_VAR.buf[MY_WRITER_VAR.s++] = a;
}
inline void wt_L(char a){
  my_putchar(a);
}
inline void wt_L(int x){
  int s=0;
  int m=0;
  char f[10];
  if(x<0){
    m=1;
    x=-x;
  }
  while(x){
    f[s++]=x%10;
    x/=10;
  }
  if(!s){
    f[s++]=0;
  }
  if(m){
    my_putchar('-');
  }
  while(s--){
    my_putchar(f[s]+'0');
  }
}
template<class T, class U> inline T GCD_L(T a, U b){
  T r;
  while(b){
    r=a;
    a=b;
    b=r%a;
  }
  return a;
}
unsigned long long HashMap_ullP_L[4];
template<class KEY, class VAL> struct HashMap{
  char*used;
  KEY*key;
  VAL*val;
  int mem;
  int n;
  int mask;
  int init_flag;
  VAL init_val;
  HashMap(){
    mem = 0;
    init_flag = 0;
  }
  ~HashMap(){
    free();
  }
  void expand(int nn){
    if(mem >= nn){
      return;
    }
    if(mem){
      free();
    }
    mem = nn;
    used = new char[nn];
    key = new KEY[nn];
    val = new VAL[nn];
  }
  void free(){
    if(mem){
      mem = 0;
      delete[] used;
      delete[] key;
      delete[] val;
    }
  }
  void init(int nn){
    int i;
    n = 1;
    nn = nn + (nn + 1) / 2;
    while(n < nn){
      n *= 2;
    }
    mask = n - 1;
    expand(n);
    for(i=(0);i<(n);i++){
      used[i] = 0;
    }
    init_flag = 0;
  }
  void init(int nn, VAL ini){
    int i;
    n = 1;
    nn = nn + (nn + 1) / 2;
    while(n < nn){
      n *= 2;
    }
    mask = n - 1;
    expand(n);
    for(i=(0);i<(n);i++){
      used[i] = 0;
    }
    init_flag = 1;
    init_val = ini;
  }
  inline int getHash(const int a){
    unsigned long long d = a;
    d = (((d * HashMap_ullP_L[0]) >> 32) * HashMap_ullP_L[1]) & mask;
    return d;
  }
  inline int getHash(const unsigned a){
    unsigned long long d = a;
    d = (((d * HashMap_ullP_L[0]) >> 32) * HashMap_ullP_L[1]) & mask;
    return d;
  }
  inline int getHash(const long long a){
    unsigned long long d = a;
    d = (((((d * HashMap_ullP_L[0]) >> 32) * HashMap_ullP_L[1]) >> 32) * HashMap_ullP_L[2]) & mask;
    return d;
  }
  inline int getHash(const unsigned long long a){
    unsigned long long d = a;
    d = (((((d * HashMap_ullP_L[0]) >> 32) * HashMap_ullP_L[1]) >> 32) * HashMap_ullP_L[2]) & mask;
    return d;
  }
  inline int getHash(const pair<int,int> a){
    unsigned long long d = (((unsigned long long)a.first) << 32) + ((unsigned long long)a.second);
    d = (((((d * HashMap_ullP_L[0]) >> 32) * HashMap_ullP_L[1]) >> 32) * HashMap_ullP_L[2]) & mask;
    return d;
  }
  inline VAL& operator[](const KEY a){
    int k = getHash(a);
    for(;;){
      if(used[k]==1 && key[k]==a){
        break;
      }
      if(used[k]==0){
        used[k] = 1;
        key[k] = a;
        if(init_flag){
          val[k] = init_val;
        }
        break;
      }
      k = (k+1) & mask;
    }
    return val[k];
  }
  inline bool exist(const KEY a){
    int k = getHash(a);
    for(;;){
      if(used[k]==1 && key[k]==a){
        return true;
      }
      if(used[k]==0){
        break;
      }
      k = (k+1) & mask;
    }
    return false;
  }
  template<class S> inline bool exist(const KEY a, S &res){
    int k = getHash(a);
    for(;;){
      if(used[k]==1 && key[k]==a){
        res = val[k];
        return true;
      }
      if(used[k]==0){
        break;
      }
      k = (k+1) & mask;
    }
    return false;
  }
}
;
int N;
char S[500000+2];
int x;
int y;
int a;
int b;
int res[500000+2];
HashMap<pair<int,int>,int> hs;
int main(){
  int Lj4PdHRW;
  {
    int i;
    int j;
    int k;
    Rand rnd;
    for(i=(0);i<(20);i++){
      rnd.get(2);
    }
    for(i=(0);i<(4);i++){
      for(j=(0);j<(32);j++){
        k = rnd.get(1,62);
        HashMap_ullP_L[i] |= (1ULL << k);
      }
      HashMap_ullP_L[i] |= (1ULL << 0);
      HashMap_ullP_L[i] |= (1ULL << 63);
    }
  }
  int KL2GvlyY = rd_int();
  for(Lj4PdHRW=(0);Lj4PdHRW<(KL2GvlyY);Lj4PdHRW++){
    int i;
    rd(N);
    rd(S);
    x = y = 0;
    hs.init(N,0);
    for(i=(0);i<(N);i++){
      if(S[i]=='K'){
        x++;
      }
      else{
        y++;
      }
      auto xr20shxY = ((x)/GCD_L(x, y));
      auto WYIGIcGE = ((y)/GCD_L(x, y));
      a=xr20shxY;
      b=WYIGIcGE;
      res[i] = ++hs[{a,b}];
    }
    {
      int t_ynMSdg;
      if(N==0){
        wt_L('\n');
      }
      else{
        for(t_ynMSdg=(0);t_ynMSdg<(N-1);t_ynMSdg++){
          wt_L(res[t_ynMSdg]);
          wt_L(' ');
        }
        wt_L(res[t_ynMSdg]);
        wt_L('\n');
      }
    }
  }
  return 0;
}
// cLay version 20210607-1 [beta]

// --- original code ---
// //no-unlocked
// int N;
// char S[5d5+2];
// int x, y, a, b, res[];
// HashMap<pair<int,int>,int> hs;
// {
//   REP(rd_int()){
//     rd(N,S);
//     x = y = 0;
//     hs.init(N,0);
//     rep(i,N){
//       if[S[i]=='K',x,y]++;
//       (a,b) = (x,y) / gcd(x,y);
//       res[i] = ++hs[{a,b}];
//     }
//     wt(res(N));
//   }
// }