message: in case, it might happen 2 children have same parent id so this case of duplicate ids and findMany will break for duplicate ids.
diff: @@ -154,12 +154,25 @@ export function embeddedChildren1(model: Mongoose.Model<any>, values: Array<any>
  if (!val[m.propertyKey] || (val[m.propertyKey] instanceof Array && !val[m.propertyKey].length))
  return;
  + let idsStr = ids.map(id => id.toString());
  if (m.propertyType.isArray) {
  - ids = ids.concat(val[m.propertyKey]);
  + // get unique ids from ids array, in case of manyToOne relationship it might happen 2 children have same parent id so this case of duplicate ids
  + // for example- student has one school, so it the case of manyToOne, in student-school relationship, so considering this case, if we going to find many two students
  + // simultenously then their parent school id are same and that is the case of duplicate ids.
  + val[m.propertyKey].forEach(id => {
  + if (idsStr.indexOf(id.toString()) === -1) {
  + ids.push(id);
  + }
  + });
  }
  else {
  + // get unique ids from ids array, in case of manyToOne relationship it might happen 2 children have same parent id so this case of duplicate ids
  + // for example- student has one school, so it the case of manyToOne, in student-school relationship, so considering this case, if we going to find many two students
  + // simultenously then their parent school id are same and that is the case of duplicate ids.
  + if (idsStr.indexOf(val[m.propertyKey].toString()) === -1) {
  ids.push(val[m.propertyKey]);
  }
  + }
  });
  if (ids.length == 0)
  return;
  