#include <bits/stdc++.h>   using namespace std;   const int NR = 4e6 + 5;   int n;   long long P = 51123987;   char s[NR];   int d[NR];   long long f[NR], g[NR];   int main() {     scanf( %d , &n);     scanf( %s , s + 1);     for (int i = n; i >= 1; i--) {       s[i * 2] = s[i];       s[i * 2 - 1] =  # ;     }     s[n * 2 + 1] =  # ;     n = n * 2 + 1;     int p = 0, maxr = 0;     for (int i = 1; i <= n; i++) {       if (i <= maxr) d[i] = min(d[p + maxr - i], maxr - i + 1);       while (i - d[i] >= 1 && i + d[i] <= n && s[i - d[i]] == s[i + d[i]]) d[i]++;       if (i + d[i] - 1 > maxr) {         p = i - d[i] + 1;         maxr = i + d[i] - 1;       }       if (s[i] ==  #  && d[i] == 1) continue;       f[(i - d[i] + 2) / 2]++;       f[i / 2 + 1]--;       if (s[i] ==  # )         g[i / 2 + 1]++;       else         g[i / 2]++;       g[(i + d[i] - 1) / 2 + 1]--;     }     n = (n - 1) / 2;     for (int i = 1; i <= n; i++) {       f[i] += f[i - 1];       f[i] %= P;       g[i] += g[i - 1];       g[i] %= P;     }     long long sum1 = 0, sum2 = 0, ans = 0;     for (int i = 1; i <= n; i++) {       sum1 += f[i];       sum1 %= P;       ans += f[i] * sum2;       ans %= P;       sum2 += g[i];       sum2 %= P;     }     sum1 = sum1 * (sum1 - 1) / 2 % P;     printf( %lld n , (sum1 - ans + P) % P);     return 0;   }   