suffixArray function used in skew algorithm

``I was studying about suffix array and came to know about skew algorithm.Here is the source code of it

 inline bool leq(int a1, int a2, int b1, int b2) // lexicographic order
{return(a1 < b1 || a1 == b1 && a2 <= b2); } // for pairs
inline bool leq(int a1, int a2, int a3, int b1, int b2, int b3)
{return(a1 < b1 || a1 == b1 && leq(a2,a3, b2,b3)); } // and triples
// stably sort a[0..n-1] to b[0..n-1] with keys in 0..K from r
static void radixPass(int* a, int* b, int* r, int n, int K)
{// count occurrences
    int* c = new int[K + 1]; // counter array
    for (int i = 0; i <= K; i++) c[i] = 0; // reset counters
    for (int i = 0; i < n; i++) c[r[a[i]]]++; // count occurrences


     for (int i = 0, sum = 0; i <= K; i++) // exclusive prefix sums
    {int t = c[i]; c[i] = sum; sum += t; }

for (int i = 0; i < n; i++) b[c[r[a[i]]]++] = a[i]; // sort
delete [] c;
}
// find the suffix array SA of s[0..n-1] in {1..K}ˆn
// require s[n]=s[n+1]=s[n+2]=0, n>=2
void suffixArray(int* s, int* SA, int n, int K) {
int n0=(n+2)/3, n1=(n+1)/3, n2=n/3, n02=n0+n2;
int* s12 = new int[n02 + 3]; s12[n02]= s12[n02+1]= s12[n02+2]=0;
int* SA12 = new int[n02 + 3]; SA12[n02]=SA12[n02+1]=SA12[n02+2]=0;
int* s0 = new int[n0];
int* SA0 = new int[n0];
// generate positions of mod 1 and mod 2 suffixes
// the "+(n0-n1)" adds a dummy mod 1 suffix if n%3 == 1
for (int i=0, j=0; i < n+(n0-n1); i++) if (i%3 != 0) s12[j++] = i;
// lsb radix sort the mod 1 and mod 2 triples
radixPass(s12 , SA12, s+2, n02, K);
radixPass(SA12, s12 , s+1, n02, K);
radixPass(s12 , SA12, s , n02, K);
// find lexicographic names of triples
int name = 0, c0 = -1, c1 = -1, c2 = -1;
for (int i = 0; i < n02; i++) {
if (s[SA12[i]] != c0 || s[SA12[i]+1] != c1 || s[SA12[i]+2] != c2)
{name++; c0 = s[SA12[i]]; c1 = s[SA12[i]+1]; c2 = s[SA12[i]+2]; }
if (SA12[i] % 3 == 1) { s12[SA12[i]/3] = name; } // left half
else {s12[SA12[i]/3 + n0] = name; } // right half
}
// recurse if names are not yet unique
if (name < n02) {
suffixArray(s12, SA12, n02, name);
// store unique names in s12 using the suffix array
for (int i = 0; i < n02; i++) s12[SA12[i]] = i + 1;
} else // generate the suffix array of s12 directly
for (int i = 0; i < n02; i++) SA12[s12[i] - 1] = i;
// stably sort the mod 0 suffixes from SA12 by their first character
for (int i=0, j=0; i < n02; i++) if (SA12[i] < n0) s0[j++] = 3*SA12[i];
radixPass(s0, SA0, s, n0, K);
// merge sorted SA0 suffixes and sorted SA12 suffixes
for (int p=0, t=n0-n1, k=0; k < n; k++) {
#define GetI() (SA12[t] < n0 ? SA12[t]*3+1: (SA12[t] - n0) * 3 + 2)
int i = GetI(); // pos of current offset 12 suffix
int j = SA0[p]; // pos of current offset 0 suffix
if (SA12[t] < n0 ? // different compares for mod 1 and mod 2 suffixes
leq(s[i], s12[SA12[t] + n0], s[j], s12[j/3]) :Simple Linear Work Suffix Array Construction 955
leq(s[i],s[i+1],s12[SA12[t]-n0+1], s[j],s[j+1],s12[j/3+n0]))
{// suffix from SA12 is smaller
SA[k] = i; t++;
if (t == n02) // done --- only SA0 suffixes left
for (k++; p < n0; p++, k++) SA[k] = SA0[p];
} else {// suffix from SA0 is smaller
SA[k] = j; p++;
if (p == n0) // done --- only SA12 suffixes left
for (k++; t < n02; t++, k++) SA[k] = GetI();
}
}
delete [] s12; delete [] SA12; delete [] SA0; delete [] s0;
}

I am little confused about the forth argument 'K' of this function

void suffixArray(int* s, int* SA, int n, int K)

What will be the value of it? It will be very helpful if someone can help me to figure this out.Thanks in advance.