Mongodb 执行聚合以消除重复并获得唯一计数

我有像下面这样的Mongo documents，想要执行聚合，以找到每个唯一的联系人的错误代码计数.

{
  "_id": {
    "$oid": ""
  }, 
  "campId": "61baef7817cd8ff66518", //camp1
  "contactId": "61aa6fbf77490b0007714273", // contact 1
  "title": "Happy Holidays!",
  "communicationType": "EMAIL", 
  "contactedOnTime": {
    "$numberLong": "1695182032" // AT TIME1
  },  
  "communicationValidationError": "EMAIL_ADDRESS_NOT_PRESENT"
}

{
    "_id": {
        "$oid": ""
    },
    "campId": "61baef7817cd8ff66518", //camp1
    "contactId": "61aa6fbf77490b0007714273", // contact1
    "title": "Happy Holidays!",
    "communicationType": "EMAIL",  
    "contactedOnTime": {
        "$numberLong": "1695182074" // AT TIME2 
    },
    "communicationValidationError": "EMAIL_ADDRESS_NOT_PRESENT"
}
{
    "_id": {
        "$oid": ""
    },
    "campId": "61baef7817cd8ff66518", // camp1
    "contactId": "61aa6fbf77490b0007714274", // contact2
    "title": "Happy Holidays!",
    "communicationType": "EMAIL",
    "contactedOnTime": {
        "$numberLong": "1695182059" 
    },
    "communicationValidationError": "EMAIL_BOUNCED"
}

我在下面试过了，但是这并不能消除Camp的重复联系人，并且显示为2而不是1.我需要为唯一的联系人 Select 最新的Communication ValidationError，并将其作为campID的Communication ValidationError的totalcounts%.

db.myCollection.aggregate([
    { $project: { _id: 0, campId: 1, contactId: 1, communicationValidationError: 1 } },
    { $sort: { "contactedOnTime.numberLong":-1}},
    { $match: { campId: '61baef7817cd8ff66518' } },
    { $group: { _id: { communicationValidationError: '$communicationValidationError' }, totalErrors: { $sum: 1 } } }
]).pretty()