diff --git a/src/main/java/g2301_2400/s2322_minimum_score_after_removals_on_a_tree/readme.md b/src/main/java/g2301_2400/s2322_minimum_score_after_removals_on_a_tree/readme.md
index 48f8a51f0..4e261bd29 100644
--- a/src/main/java/g2301_2400/s2322_minimum_score_after_removals_on_a_tree/readme.md
+++ b/src/main/java/g2301_2400/s2322_minimum_score_after_removals_on_a_tree/readme.md
@@ -23,7 +23,15 @@ Return _the **minimum** score of any possible pair of edge removals on the given
 
 **Output:** 9
 
-**Explanation:** The diagram above shows a way to make a pair of removals. - The 1<sup>st</sup> component has nodes [1,3,4] with values [5,4,11]. Its XOR value is 5 ^ 4 ^ 11 = 10. - The 2<sup>nd</sup> component has node [0] with value [1]. Its XOR value is 1 = 1. - The 3<sup>rd</sup> component has node [2] with value [5]. Its XOR value is 5 = 5. The score is the difference between the largest and smallest XOR value which is 10 - 1 = 9. It can be shown that no other pair of removals will obtain a smaller score than 9. 
+**Explanation:** The diagram above shows a way to make a pair of removals.
+
+- The 1<sup>st</sup> component has nodes [1,3,4] with values [5,4,11]. Its XOR value is 5 ^ 4 ^ 11 = 10.
+
+- The 2<sup>nd</sup> component has node [0] with value [1]. Its XOR value is 1 = 1.
+
+- The 3<sup>rd</sup> component has node [2] with value [5]. Its XOR value is 5 = 5.
+
+The score is the difference between the largest and smallest XOR value which is 10 - 1 = 9. It can be shown that no other pair of removals will obtain a smaller score than 9. 
 
 **Example 2:**
 
@@ -33,7 +41,15 @@ Return _the **minimum** score of any possible pair of edge removals on the given
 
 **Output:** 0
 
-**Explanation:** The diagram above shows a way to make a pair of removals. - The 1<sup>st</sup> component has nodes [3,4] with values [4,4]. Its XOR value is 4 ^ 4 = 0. - The 2<sup>nd</sup> component has nodes [1,0] with values [5,5]. Its XOR value is 5 ^ 5 = 0. - The 3<sup>rd</sup> component has nodes [2,5] with values [2,2]. Its XOR value is 2 ^ 2 = 0. The score is the difference between the largest and smallest XOR value which is 0 - 0 = 0. We cannot obtain a smaller score than 0. 
+**Explanation:** The diagram above shows a way to make a pair of removals.
+
+- The 1<sup>st</sup> component has nodes [3,4] with values [4,4]. Its XOR value is 4 ^ 4 = 0.
+
+- The 2<sup>nd</sup> component has nodes [1,0] with values [5,5]. Its XOR value is 5 ^ 5 = 0.
+
+- The 3<sup>rd</sup> component has nodes [2,5] with values [2,2]. Its XOR value is 2 ^ 2 = 0.
+
+The score is the difference between the largest and smallest XOR value which is 0 - 0 = 0. We cannot obtain a smaller score than 0. 
 
 **Constraints:**