#include <bits/stdc++.h>   using namespace std;   const int MOD = 1000000007;   int N, dp[5000][5001], cumlcp[5000][5000];   vector<int> sa, ra, rsa, lcp;   string s;   void saInit() {     int l = (int)s.size();     sa.resize(l);     iota(sa.begin(), sa.end(), 0);     ra.assign(s.begin(), s.end());     for (int k = 1; k < l; k *= 2) {       sort(sa.begin(), sa.end(), [&](int a, int b) {         if (ra[a] != ra[b]) return ra[a] < ra[b];         int ak = a + k < l ? ra[a + k] : -1;         int bk = b + k < l ? ra[b + k] : -1;         return ak < bk;       });       vector<int> ra2(l);       int x = 0;       for (int i = 1; i < l; i++) {         if (ra[sa[i]] != ra[sa[i - 1]] || sa[i - 1] + k >= l ||             ra[sa[i] + k] != ra[sa[i - 1] + k])           x++;         ra2[sa[i]] = x;       }       ra = ra2;     }     rsa.resize(l);     for (int i = 0; i < int(l); i++) rsa[sa[i]] = i;   }   void saLCP() {     int l = (int)s.size();     lcp.resize(l);     vector<int> p(l);     for (int i = 0; i < l; i++) {       p[sa[i]] = (i) ? sa[i - 1] : -1;     }     int x = 0;     for (int i = 0; i < l; i++) {       while (p[i] != -1 && p[i] + x < l && s[i + x] == s[p[i] + x] &&              s[i + x] !=  $ )         x++;       lcp[rsa[i]] = x;       if (x) x--;     }     for (int i = 0; i < int(l); i++) {       cumlcp[i][i] = INT_MAX;       for (int j = i + 1; j < int(l); j++)         cumlcp[i][j] = min(cumlcp[i][j - 1], lcp[j]);     }   }   bool ism(int a, int b, int n) {     int as = a - n + 1, bs = b - n + 1;     if (as < 0) return false;     if (rsa[bs] <= rsa[as]) return false;     return cumlcp[rsa[as]][rsa[bs]] < n;   }   int main() {     cin >> N >> s;     saInit();     saLCP();     for (int i = 0; i < int(N); i++)       for (int j = 1; j < int(5001); j++) {         int& ans = dp[i][j];         ans = dp[i][j - 1];         if (j > i + 1) {           continue;         } else if (s[i - j + 1] ==  0 ) {         } else if (j == i + 1) {           ans = (ans + 1) % MOD;         } else if (ism(i - j, i, j)) {           ans = (ans + dp[i - j][j]) % MOD;         } else {           ans = (ans + dp[i - j][j - 1]) % MOD;         }       }     cout << dp[N - 1][N] << endl;   }   