diff --git a/src/string/ac-automaton.hpp b/src/string/ac-automaton.hpp
index d2a010c..4ed4538 100644
--- a/src/string/ac-automaton.hpp
+++ b/src/string/ac-automaton.hpp
@@ -4,8 +4,10 @@
 #include <queue>
 /** Modified from:
  * https://github.com/kth-competitive-programming/kactl/blob/master/content/strings/AhoCorasick.h
- * Try to handdle duplicated patterns beforehand, otherwise change 'end' to
- * vector; empty patterns are not allowed. Time: construction takes $O(26N)$,
+ * If there's no duplicated patterns, just call the constructor, otherwise handle it beforehand
+ * by yourself, or use the return value of insert
+ * empty patterns are not allowed.
+ * Time: construction takes $O(26N)$,
  * where $N =$ sum of length of patterns. find(x) is $O(N)$, where N = length of
  * x. findAll is $O(N+M)$ where M is number of occurrence of all pattern (up to N*sqrt(N)) */
 
@@ -14,8 +16,7 @@ struct AhoCorasick {
     struct Node {
         // back: failure link, points to longest suffix that is in the trie.
         // end: longest pattern that ends here, is -1 if no patten ends here.
-        // nmatches: number of (patterns that is a suffix of current
-        // node)/(duplicated patterns), depends on needs.
+        // nmatches: number of patterns that is a suffix of current node
         // output: output link, points to the longest pattern that is a suffix
         // of current node
         int back, end = -1, nmatches = 0, output = -1;
@@ -33,7 +34,9 @@ struct AhoCorasick {
         build();
     }
 
-    void insert(const std::string &s, int j) { // j: id of string s
+    // returns -1 if there's no duplicated pattern already in the trie
+    // returns the id of the duplicated pattern otherwise
+    int insert(const std::string &s, int j) { // j: id of string s
         assert(!s.empty());
         int n = 0;
         for (char c : s) {
@@ -43,13 +46,19 @@ struct AhoCorasick {
             }
             n = N[n].next[c - first];
         }
+        if (N[n].end != -1) {
+            return N[n].end;
+        }
         N[n].end = j;
         N[n].nmatches++;
+        return -1;
     }
 
     void build() {
+        // adds a dummy node so the root node can be correctly handled
         N[0].back = (int)N.size();
         N.emplace_back(0);
+
         std::queue<int> q;
         q.push(0);
         while (!q.empty()) {
@@ -64,14 +73,13 @@ struct AhoCorasick {
                     // if prev is an end node, then set output to prev node,
                     // otherwise set to output link of prev node
                     N[v].output = N[fail].end == -1 ? N[fail].output : fail;
-                    // if we don't want to distinguish info of patterns that is
-                    // a suffix of current node, we can add info to the next
-                    // node like this: nxt.nmatches+=N[pnx].nmatches;
+                    N[v].nmatches += N[fail].nmatches;
                     q.push(v);
                 }
             }
         }
     }
+
     // for each position, finds the longest pattern that ends here
     std::vector<int> find(const std::string &text) {
         int len = (int)text.size();
@@ -83,14 +91,17 @@ struct AhoCorasick {
         }
         return res;
     }
-    // for each position, finds the all that ends here
+
+    // for each position, finds all patterns that ends here
     std::vector<std::vector<int>> find_all(const std::string &text) {
         int len = (int)text.size();
         std::vector<std::vector<int>> res(len);
         int n = 0;
         for (int i = 0; i < len; i++) {
             n = N[n].next[text[i] - first];
-            res[i].push_back(N[n].end);
+            if (N[n].end != -1) {
+                res[i].push_back(N[n].end);
+            }
             for (int ind = N[n].output; ind != -1; ind = N[ind].output) {
                 assert(N[ind].end != -1);
                 res[i].push_back(N[ind].end);
@@ -99,8 +110,9 @@ struct AhoCorasick {
         return res;
     }
 
-    std::vector<int> find_cnt(const std::string& text, int n) {
-        std::vector<int> cnt(n);
+    // finds the number of occurrence of each pattern
+    std::vector<int> find_cnt(const std::string& text, int num_of_patterns) {
+        std::vector<int> cnt(num_of_patterns);
         int p = 0;
         for (auto c : text) {
             p = N[p].next[c - first];